View Full Version : Hitchcraft for hammock tensioning
Hi, I.. I mean a totally obsessed friend of mine... was searching through hammock videos on Youtube when this video came up
MFHEQOqX7vQ
basically what appears to be a simple to use pulley/tie off system
I did a bit of leg work and have the link to the hitchcraft site
http://www.hitchcraft.net/
I thought it might be of interest to hangers as a possible alternative to the Hennessey whipping or the strap/rings method. I don't use a HH so have no real reason to get these I just thought somebody might find these useful.
Two sizes available. I'm not sure what diameter rope is used on a HH but I assume its the smaller version so 250 lbf (though hammockers are looking at two points of support halving the forces?!?).
Mini RopeTie
weight: 0.9 oz (26 g)
safe working load: 250 lbf (1.1 kN)
rope diameter: .125"-.250" / 3mm-6mm
$11.95 /pr
Monster RopeTie
weight: 2.9 oz (82 g)
safe working load: 500lbf (2.2 kN)
rope diameter: .250" - .438" / 6mm-11mm
$17.95
2 mini + 2 monster $27.95
Drop
FanaticFringer
05-30-2007, 20:44
Here's a thread about it:
www.hammockforums.net/forum/showthread.php?t=980&highlight=monster+rope
Doh, sorry, thought I new all the threads by heart now.
feel free to lock/delete mods
FanaticFringer
05-30-2007, 20:50
Dont feel bad. I did the same thing yesterday about the Tom Hennessy response.:(
hitchman
05-31-2007, 11:45
Hi!
I saw the threads that mention my product, the Hitchcraft RopeTie, and I would like to address some of the comments.
This product was the result of a couple of years development. The jam cleat that makes it adjustable does not damage the rope. The edges are rounded and smooth. This type of device is commonly found in sailboats and windsurfer rope handling gear.
Personally I have a Hennessey Hammock with a 1/4" rope. I understand there are other models with smaller spectra line. The Mini version will work with 1/8" to 1/4" line.
Thanks to those who showed interest in my product. I'd be happy to respond personally to any questions.
Cheers!
Hitchman, will the mini's hold the load required to keep a Hennessey hammock off the ground. We are talking forces over 6 to 7 hundred pounds or more?
hitchman
05-31-2007, 12:55
By my calculation the load on each end rope is about the same as the person's weight. I assume an angle of 30 degrees when loaded, which by my experience requires a pretty tight rope.
The safe load on the Mini RopeTie is 250 lbs. The breaking load is more than twice that. Given that, I would recommend the Monster for people who weigh close to 200 lbs. Backpackers will have to weigh the pros and cons of carrying 4 oz of the Monster pair versus the Mini. Alternatively, if you carry only one Monster, that will give you quick adjustment on one end, which is all you really need.
However, if you carry some rope, it would be advisable to carry the size that also works with that rope.
Thanks for your question.
Miguel
blackbishop351
05-31-2007, 13:28
By my calculation the load on each end rope is about the same as the person's weight. I assume an angle of 30 degrees when loaded, which by my experience requires a pretty tight rope.
You might want to recheck your math. Just an estimate here, but a 200-lb. person should exert around 600 lbs. of tension on a suspension, with a 30* sag angle.
hitchman
05-31-2007, 13:56
Here is my math: Sin of 30 is 0.5, meaning that for a 30 degree angle the vertical component is half of the tension on the rope. Each end carries half of the weight. So for a 200 lb weight, or any weight for that matter, the rope tension on each end is the same as the weight (multiplied by 0.5 and divided by 2).
I agree that the conservative way to go is with the Monster, but many back packers watch their gear weight very closely.
I weigh 180 lbs and feel quite safe with the Mini, but it is important to note that this device makes tensioning very easy, and people may be inclined to tighten the rope more than they would otherwise. With a tighter rope the angle will get smaller and the load on the rope and the hitch will increase.
blackbishop351
05-31-2007, 14:29
Here is my math: Sin of 30 is 0.5, meaning that for a 30 degree angle the vertical component is half of the tension on the rope. Each end carries half of the weight. So for a 200 lb weight, or any weight for that matter, the rope tension on each end is the same as the weight (multiplied by 0.5 and divided by 2).
I agree that the conservative way to go is with the Monster, but many back packers watch their gear weight very closely.
I weigh 180 lbs and feel quite safe with the Mini, but it is important to note that this device makes tensioning very easy, and people may be inclined to tighten the rope more than they would otherwise. With a tighter rope the angle will get smaller and the load on the rope and the hitch will increase.
Yeah so apparently I automatically can't do geometry in my head after graduation...thanks! :p
GrizzlyAdams
05-31-2007, 15:26
Here is my math: Sin of 30 is 0.5, meaning that for a 30 degree angle the vertical component is half of the tension on the rope. Each end carries half of the weight. So for a 200 lb weight, or any weight for that matter, the rope tension on each end is the same as the weight (multiplied by 0.5 and divided by 2).
W = weight
W*0.5/2 = W/4
not W.
??
The thing to worry about is dynamic tension, not static, me thinks. Getting mass times acceleration involved, and all that.
Grizz
hitchman
05-31-2007, 15:38
The 0.5 factor is multiplied not divided. The tension on the rope is the resultant from the vertical component (weight) and the horizontal component (pulling the trees together). So at 30 deg the tension on the rope is twice the vertical load (half the weight on each end).
I wouldn't think the dynamic component is that great unless you are in earthquake country. Otherwise let me point out that the rope is rated at the breaking point (for new undamaged rope). Rope accessories are generally rated at the recommended work load, which is around 1/3 of the breaking load. :D
FanaticFringer
05-31-2007, 17:18
I hate math.:D
blackbishop351
05-31-2007, 20:16
Math is hard.
I have that on a t-shirt...nobody gets the joke, though :( :rolleyes:
GrizzlyAdams
05-31-2007, 21:45
There are two kinds of math : easy math, and hard math.
Easy math is the kind you understand, hard math is the kind you don't.
Also a joke that most don't get, or at least don't think is funny.
I've got a degree in the stuff, my wife's joke is that I double majored in math, and more math. However my courses ran to the crystal tower variety, and not the useful stuff like elementary engineering.
Not to belabor the point, but if you take half the weight (W/2) and if
The 0.5 factor is multiplied not divided
then that's (W/2)*0.5. But these are just words...I now understand the physics of the situation by your description
The tension on the rope is the resultant from the vertical component (weight) and the horizontal component (pulling the trees together).
So I gather that the way to think about this is that the known force is W/2, understood to be the vertical projection of the (unknown) total force T on the rope. The ratio of the vertical projection to the total component (hypotenus) is the sine of the angle theta between the hypotenus and adjacent side, so sine(theta) = (W/2)/T. With sine( pi/6 ) = 0.5 we solve for T in 0.5 = (W/2)/T, which is T = (W/2)/0.5 = W.
When you say that the 0.5 factor is multiplied and not divided, I guess you mean 0.5*T = (W/2).
so all is clear now.
Grizz the occasionally obtuse
hitchman
05-31-2007, 22:30
Couldn't have put it better myself!
angrysparrow
05-31-2007, 22:33
There are two kinds of math : easy math, and hard math.
Easy math is the kind you understand, hard math is the kind you don't.
I personally always thought 'hard math' was the kind that Newton, Bernoulli, Leibniz, Euclid, and Einstein struggled with. :rolleyes:
so all is clear now.
Yes, the outlined principle is clear enough. It doesn't make me any less inclined to want my hammock ridiculously over-engineered, though. My backside, and my ego, don't like taking falls. ;)
For those of you who do math so well, I have but one thing to say. I hate you. :mad: :eek: Just kidding, :p thanks for the science to back up the hanging, I appreciate it. :D
There are two kinds of math : easy math, and hard math.
Easy math is the kind you understand, hard math is the kind you don't.
Also a joke that most don't get, or at least don't think is funny.
I've got a degree in the stuff, my wife's joke is that I double majored in math, and more math. However my courses ran to the crystal tower variety, and not the useful stuff like elementary engineering.
Not to belabor the point, but if you take half the weight (W/2) and if
then that's (W/2)*0.5. But these are just words...I now understand the physics of the situation by your description
So I gather that the way to think about this is that the known force is W/2, understood to be the vertical projection of the (unknown) total force T on the rope. The ratio of the vertical projection to the total component (hypotenus) is the sine of the angle theta between the hypotenus and adjacent side, so sine(theta) = (W/2)/T. With sine( pi/6 ) = 0.5 we solve for T in 0.5 = (W/2)/T, which is T = (W/2)/0.5 = W.
When you say that the 0.5 factor is multiplied and not divided, I guess you mean 0.5*T = (W/2).
so all is clear now.
Grizz the occasionally obtuse
Hey now show some respect.
Hitchman, you have a good looking product. I broke and got dropped using 550 lbs rated paracord. It broke in the middle and not at the knot. It was at a decent angle, 30-45 deg. For me it is nothing rating below 1000lbs.
"multiplied by 0.5" = "divided by 2"
Maybe the confusion was stating the same thing two different ways.
GrizzlyAdams
06-01-2007, 10:44
Hey now show some respect.
hey, another Atkins-Knofler fan!
http://www.hammockforums.net/gallery/misc.php?do=downloadfile&i=929
Grizz the obsequious
hitchman
06-01-2007, 11:08
Hey now show some respect.
Hitchman, you have a good looking product. I broke and got dropped using 550 lbs rated paracord. It broke in the middle and not at the knot. It was at a decent angle, 30-45 deg. For me it is nothing rating below 1000lbs.
Engineer, the rating on ropes has a totally different meaning than the rating on most products. If you read rope manufacturers specs they tell you what the test breaking strength of brand new rope is. Then they recommend you to use it at a maximum of 20-25% of that and use your judgement to be even more conservative when loss of life (or sleep) is at stake. Clearly the cord you used was degraded. Not too long ago a well known climber fell to his death because he was using a worn out harness that broke. Those things are designed with a factor of safety of 10 (or something close to that, I'm improvising a little).
When you look at equipment for rope handling (pulleys, etc) you will find that the safe working load (SWL) is much less than what the rope sized for it can take. In general it is close to 25% of the average rope strength. The point I'm trying to make is that using a 550lb rope is equivalent to using a device rated for 137lbs (SWL).
Engineer, the rating on ropes has a totally different meaning than the rating on most products. If you read rope manufacturers specs they tell you what the test breaking strength of brand new rope is. Then they recommend you to use it at a maximum of 20-25% of that and use your judgement to be even more conservative when loss of life (or sleep) is at stake. Clearly the cord you used was degraded. Not too long ago a well known climber fell to his death because he was using a worn out harness that broke. Those things are designed with a factor of safety of 10 (or something close to that, I'm improvising a little).
When you look at equipment for rope handling (pulleys, etc) you will find that the safe working load (SWL) is much less than what the rope sized for it can take. In general it is close to 25% of the average rope strength. The point I'm trying to make is that using a 550lb rope is equivalent to using a device rated for 137lbs (SWL).
I am not an expert or even well read on this, so don't take this as me arguing with you.
When I broke paracord (twice) I cut a fresh piece off the roll and tied it on the hammock. Both times it broke the first time I used it.
If I was using a rope rated to 550 lbs, shouldn't it hold on the first use at least that?
From what little I understand on pulleys and racketing straps, isn't the rating governed by the pulley or racket itself and not the rope?
In the end though I am probibly the most conservative person here in terms of material strengh. I am using 7/54 armsteel sprectra rating around 1400lbs, rings rating over 2000 lbs, biner's rated over 2000 lbs, and webbing rated over 1400 lbs. In the end I think the hitches I use to attach the hammock support ropes to the hammock and rings are the weakest link.
hitchman
06-01-2007, 14:12
I'm learning as much from this conversation as anybody so your views are welcome. I don't see htis as an argument.
Anyway, I'm surprised that the 550lbs rated cord would break. Under the conditions you describe it doesn't seem like it had anymore than half that load. I would confront the manufacturer with that fact.
All the ratings you mention are probably breaking strength, and I can tell you that the Hitchcraft Monster tie failed at 1800lbs in tests, so that puts it in the same ball park. I just think that when you see a pulley, for instance, rated at 500lbs (SWL) and you use it with a 2000lbs cord, that doesn't necesarily mean the pulley is the weakest link.
hitchman - I think you are overlooking the horizontal forces. So far you have considered only the vertical force, the weight, and seem to feel that the horizontal forces are not worth mentioning. But most, maybe all, here that use a structural ridgeline, pull that ridgeline very tight when hanging. So the horizontal forces are not insignificant.
I personally pulled a brand new hook of 1/4" diameter steel out into a straight line the first time I hung a hammock with a structural ridgeline. Pulled the ridgeline tight, got in the hammock, was just relaxing and got dropped to the ground. But it took a hell of a lot more than my weight to pull that 1/4" steel hook into a straight line. So I think you are not really considering all of the forces at play here.
blackbishop351
06-01-2007, 16:19
hitchman - I think you are overlooking the horizontal forces. So far you have considered only the vertical force, the weight, and seem to feel that the horizontal forces are not worth mentioning. But most, maybe all, here that use a structural ridgeline, pull that ridgeline very tight when hanging. So the horizontal forces are not insignificant.
I personally pulled a brand new hook of 1/4" diameter steel out into a straight line the first time I hung a hammock with a structural ridgeline. Pulled the ridgeline tight, got in the hammock, was just relaxing and got dropped to the ground. But it took a hell of a lot more than my weight to pull that 1/4" steel hook into a straight line. So I think you are not really considering all of the forces at play here.
He isn't diagramming a hammock with a structural ridgeline, and he's correct as to the suspension forces for one without a ridge.
GrizzlyAdams
06-01-2007, 16:24
hitchman - I think you are overlooking the horizontal forces. So far you have considered only the vertical force, the weight, and seem to feel that the horizontal forces are not worth mentioning. But most, maybe all, here that use a structural ridgeline, pull that ridgeline very tight when hanging. So the horizontal forces are not insignificant.
Question for the engineering crowd here. Suppose when tightening up the ridgeline we pull it at P lbs. Then a vertical weight W plunks down in the hammock, forming an angle theta between the rope and the horizontal. What's the formula for the stress on the rope? Thanks.
Grizz
blackbishop351
06-01-2007, 16:38
The same as if you were considering a hammock without a ridgeline - just consider the hammock/ridge/user system as a solid mass.
That's only considering the suspension between the ridgeline attachment point and the tree, though...between the hammock and the attachment point gets more interesting. Then you get to figure the tension in the ridgeline, too. I'm not sure how it works out right off the top of my head...I'll play with it when I get home from work later, though.
He isn't diagramming a hammock with a structural ridgeline, and he's correct as to the suspension forces for one without a ridge.
Agreed, but he's said he's using a Hennessy, and the Hitchcraft site depicts a Hennessy being hung. Also, I believe that Hammock Engineer uses a structural ridgeline. Thus, I still do not think he is considering all the forces at play. If he were, then that steel hook would still be unchanged.
Just Jeff
06-01-2007, 16:42
The force vector on the rope is the hypoteneus of a triangle.
h = (.5 x user weight) / sin(support angle)
GrizzlyAdams
06-01-2007, 17:01
BB and JJ--before we create the vertical tension on the rope, I break a sweat and pull on that rope as hard as I can, and fasten it. P lbs is what I can apply on a good day. So there is that much tension on the rope before climbing into the hammock. Now I climb in and create the additional stress whose formula we know. P did not go away. So is the total tension P + ASWFWK ? Does trigonometry come sneaking back here too?
BB--yeah, I can see that the stress on the ridgeline could have some interesting geometry.
Thanks.
Grizz
blackbishop351
06-01-2007, 18:05
BB and JJ--before we create the vertical tension on the rope, I break a sweat and pull on that rope as hard as I can, and fasten it. P lbs is what I can apply on a good day. So there is that much tension on the rope before climbing into the hammock. Now I climb in and create the additional stress whose formula we know. P did not go away. So is the total tension P + ASWFWK ? Does trigonometry come sneaking back here too?
BB--yeah, I can see that the stress on the ridgeline could have some interesting geometry.
Thanks.
Grizz
The ridgeline is how that extra tension gets added in...like I said, I'll see about drawing something up later this evening.
GrizzlyAdams
06-01-2007, 19:18
The ridgeline is how that extra tension gets added in...like I said, I'll see about drawing something up later this evening.
Appreciated. If you haven't started already, here's a sketch of the simplest problem I can think of that whose solution would illustrate the principle I'm asking about.
A cord is originally horizontal, stretched to be under tension at an equivalence of P lbs (at the center, if position matters).
A weight of W lbs is attached to the center, causing the line to sag
and creating an angle theta between the line and its former horizontal position. What now is the tension on the rope?
http://www.hammockforums.net/gallery/files/4/2/3/Picture10.png
Let the mathemagic begin!
Thanks,
Grizz
OK, lets say we get this math thing all worked out. But now lets throw in a new twist. What's the failure rating at below freezing temperatures? I'm over fifty and I tend to bounce less when I'm cold!
The force vector on the rope is the hypoteneus of a triangle.
h = (.5 x user weight) / sin(support angle)
JJ - that is true only for a hammock without a ridgeline or for those with a ridgeline which isn't pulled tight when hung. For those with a ridgeline that people pull tight, that formula still ignores the horizontal force exerted in pulling the ridgeline tight.
The ridgeline is how that extra tension gets added in...l
Well - yes and no. The ridgeline doesn't create the extra force, the extra force is created when the ridgeline is pulled tight.
Appreciated. If you haven't started already, here's a sketch of the simplest problem I can think of that whose solution would illustrate the principle I'm asking about.
A cord is originally horizontal, stretched to be under tension at an equivalence of P lbs (at the center, if position matters).
A weight of W lbs is attached to the center, causing the line to sag
and creating an angle theta between the line and its former horizontal position. What now is the tension on the rope?
http://www.hammockforums.net/gallery/files/4/2/3/Picture10.png
Let the mathemagic begin!
Thanks,
Grizz
Okay, you now have the original force applied in pulling the rope/line/cord tight, plus the weight applied as indicated, plus the elastic force exerted by the fibers in the rope/line/cord used when stretched, both the stretching when the rope/line/cord is pulled tight and the additional stretching when the weight is added. Lacking a lot of experimental data, the last force is totally unknown to me. I would guess though that it can be quite high.
Also, you will note that the vertical force is the applied weight plus the weight of the rope/line/cord. That vertical force is translated into a force along the rope/line/cord which is equal to (ignoring the weight of the rope/line/cord):
W / (2 * sin(A))
where A is the angle the rope/line/cord makes with the horizontal. Now if the initial force was purely horizontal, then the force on the rope initially was:
P / (cos(B))
where B is the angle the rope made with the horizontal before adding any weight. Both of the above ignore any stretching of the rope and hence any of the elastic forces of the rope.
Now the total force, excluding the elastic force, along the rope is the sum:
(W/(2* sin(A))) + (P/cos(B))
If B is less than, say 10 deg., then the second term can be closely approximated by P, ignoring the cos(B) term. for A equal to the much used 30 deg, then the first term reduces to W as stated previously.
What the 30 deg assumption used previously blithely ignores is that for those using a structural ridgeline, it is my experience that A is very much less than 30 deg.It is more common in my experience that A is closer to 10 deg (estimated, not measured). For a value of 10 deg, the first term reduces to:
W * 2.9
to the first decimal place. For 15 deg., we have:
W * 1.9
and for 5 deg., we have:
W * 5.7
The value of A is going to very dependent on P and the elastic properties of the suspension rope and the elastic properties of the ridgeline cord. If the ridgeline and suspension cords have very little stretch, then A is going to be smaller relatively than if both are fairly elastic.
Thus, the force applied along the suspension rope for a hammock with a structural ridgeline can be much greater than the weight of the occupant. And that ignores the elastic forces which may be much greater ( a guess on my part based on my experience with bent steel hooks and without the benefit of data on the physical properties of the ropes used).
GrizzlyAdams
06-01-2007, 20:53
got it, thanks for the run down. So for a line starting at the horizonatal it's
tension = P + ASWFWK + unknown-elastic-force-that-bends-steel-S-hooks
Given how I've been cranking up the tension on the ridgeline of my HH, I'm
(a) mighty glad I've doubled up the Dyeema cord on my rings, and
(b) considering taking a step ladder along on my backpacking trips so I can just attach the lines higher, still keep my posterior off the ground, and reap the benefits of large sine(theta). :D
Grizz (who, having just crossed the magic threshold of 30 postings exits the state of probationary membership, wonders if anyone will now teach him the secret electronic handshake)
Frolicking Dino
06-01-2007, 21:06
Glad we have you mathematically-inclined fellows here to let us know it's OK to use the 200# cleats. I'll have to get some - that would make testing so much easier.
slowhike
06-01-2007, 21:19
man... i didn't know yall's going to be talkin greek on here:o
blackbishop351
06-01-2007, 21:31
Well - yes and no. The ridgeline doesn't create the extra force, the extra force is created when the ridgeline is pulled tight.
Obviously...sometimes I type too quickly for my own good. Sorry for the misuse of language there.
You're dead on with your force analysis, as far as the suspension beyond the ridgeline attachments. I'll also comment that for most purposes, the elastic forces involved can be ignored by assuming a static situation, which is perfectly reasonable. I'm not about to try and tackle a dynamic analysis :p
What I'm thinking about is the forces between the hammock and attachment points, as well as the tension on the ridgeline itself. I'll get on that sometime soon...maybe tonight, but now I've got beer on the way :p
Wow! This former history and music major is much impressed! Not only by the math skills but how well you guys did explaining it to us non-math folks!
GrizzlyAdams
06-01-2007, 22:27
man... i didn't know yall's going to be talkin greek on here:o
yeah, but it's Koine Greek, so your pastor will approve.
Grizz
slowhike
06-02-2007, 06:33
yeah, but it's Koine Greek, so your pastor will approve.
Grizz
yeah... but it's still greek to me:o
This is one of the reasons I like this site. Not only do I learn the how's and why's, but also the background.
Thanks everyone.:)
hitchman
06-02-2007, 13:56
hitchman - I think you are overlooking the horizontal forces. So far you have considered only the vertical force, the weight, and seem to feel that the horizontal forces are not worth mentioning. But most, maybe all, here that use a structural ridgeline, pull that ridgeline very tight when hanging. So the horizontal forces are not insignificant.
I personally pulled a brand new hook of 1/4" diameter steel out into a straight line the first time I hung a hammock with a structural ridgeline. Pulled the ridgeline tight, got in the hammock, was just relaxing and got dropped to the ground. But it took a hell of a lot more than my weight to pull that 1/4" steel hook into a straight line. So I think you are not really considering all of the forces at play here.
Teedee - I did ignore the ridgeline, and I see that the discussion went on. The easiest way to estimate the load is to ignore the ridgeline and look at the angle of the cord, because the effect of the ridgeline is to reduce that angle, and the angle determines the tension in function of the weight (w/2 sin A). The same formula applies. If you go to http://hitchcraft.net/thumb_7.html there is a picture that shows an angle around 30 deg. I don't know if this is typical, but if you like to tighten the ridgeline you are increasing the load very quickly as the angle changes.
Teedee - I did ignore the ridgeline, and I see that the discussion went on. The easiest way to estimate the load is to ignore the ridgeline and look at the angle of the cord, because the effect of the ridgeline is to reduce that angle, and the angle determines the tension in function of the weight (w/2 sin A). The same formula applies. If you go to http://hitchcraft.net/thumb_7.html there is a picture that shows an angle around 30 deg. I don't know if this is typical, but if you like to tighten the ridgeline you are increasing the load very quickly as the angle changes.
Ignoring the ridgeline is by far the easiest way to go but unfortunately leads to incorrect conclusions.
In one thread someone asked the members of the forum how they hung with the structural ridgeline. If I remember correctly, every responder wrote that they tighten the suspension down as tight as they reasonably could when hanging. I know I did that also when I first got my Hennessy hammock. I can no longer remember whether I picked that up from Sgt Rock's web site or from Hennessy in some manner. Just checked the Hennessy ULBA stuff sack - it advises: "Adjust hammock until centered and ridge line is level and under light tension." Also, nearly every advisement I have read on the net about hanging the Hennessy says to pull tight, sit or lay on the hammock for a few minutes and then re-tighten. So it would seem a loose ridgeline is not used by many people.
No - the formula you have given ignores the tension of the suspension prior to any weight being added. That tension in the system and the ridgeline definitely cannot be ignored. They do 3 things:
1. The tension adds to the forces on the suspension.
2. the stretching of the ridgleine when weight is added adds more force to the suspension system.
3. the ridgeline, as most people who have described their use of the structural rideline here on the forum, limits the final suspension angle to very much less than what you are assuming or apparently use yourself judging from the picture you referenced.
1. and 2. are forces for which you have not accounted and 3. greatly increases the force applied by your formula because it limits the suspension angle from what you have assumed or use.
Now ignoring the tension applied by the user when hanging is also tempting, since there is no practical way of measuring or assuring or limiting what that force is. But it most definitely cannot be ignored.
I assume you are discussing a system with a structural ridgeline since you have stated that you personally are using a Hennessy, which I further assume still has it's structural ridgeline intact, and your video depicts a Hennessy being used. As such, you must then account for the forces which you're ignoring and which cannot be ignored and you cannot assume the 30 deg suspension angle as you have been doing.
I napped in my hammock this afternoon and again "eyeballed" the suspension angle and I am pretty sure that the final suspension angle is between 5 deg and 10 deg, closer to 10 deg. That increases the forces by a factor of approximately 3 to over 5 times what you are computing with a 30 deg angle. That is significant and doesn't account for the forces in 1. and 2.
If you are discussing solely a system without a structural ridgeline, then your formula applies as is and you are probably correct in assuming something close to a 30 deg suspension angle.
Oh by the by - you may want to change the picture you referenced and the video also - both show the hammock hung with no tree huggers. Not many forum members think that is a good thing. :D
hitchman
06-04-2007, 12:42
Ignoring the ridgeline is by far the easiest way to go but unfortunately leads to incorrect conclusions.
If you are discussing solely a system without a structural ridgeline, then your formula applies as is and you are probably correct in assuming something close to a 30 deg suspension angle.
Oh by the by - you may want to change the picture you referenced and the video also - both show the hammock hung with no tree huggers. Not many forum members think that is a good thing. :D
TeeDee - I'm afraid I may not have made myself clear. When I say ignore the ridgeline I don't mean that it makes no difference. What I mean is that when you analize the forces you can cut the ridgeline out of the equation. You can look at the forces by looking at the hammock as a black box suspended by two ropes. The only thing that you need to know to calculate the forces on the rope (and rope hardware) is the angle and the load. SO with or without ridgeline, what matters is the angle of the rope when you are hanging on the hammock.
I did use a 30 degree angle, and I don't dispute that as the angle gets smaller the rope tension increases very fast. In fact I offered that on my last post. Now from many pictures and my own use I don't know that there is an advantage in tightening the hammock so much that the angle is 10 degrees. That puts huge loads on the hammock components.
I don't want to get too technical but to take this argument to conclusion I have to. By reverse engineering my HH I find that the rope has a breaking load of 1600 lbs. The manufacturer's suggested working load on a rope is 20% of that limit (from Samson Ropes literature). That tells me my hammock was designed for a maximum load of 320 lbs on the rope, which at 30 degrees means a 320 lb weight. That results in a safety factor of 1.28 for the design load of 250lbs. If you reduce the angle the factor of safety goes away. You might want to check with the manufacturer what their recommendation is, but as a product developer this is how I make my assessment.
Regarding the tree huggers, I'm with you. The picture was taken when I filmed a short instructional video about the hitchcraft tie and it was a way to keep it simple. In fact, the hitchcraft ties work better with tree huggers as long as there is enough distance between the trees.
hitchman, I may have missed this in an earlier post, but how many hangs do you have on the same set? Actually how many short hangs or over nights.
hitchman
06-04-2007, 13:34
Are you referring to the life of the product?
I don't have much data, but I have a demo that has been used for 100-150 cycles and the teeth start to show some wear marks. I would estimate that they still have at least 70-80% of their life left and would have to be replaced after 500-1000 cycles of hard use.
I would expect that they would last for the life of the hammock and still be fully functional when the hammock components (seams, ropes) start to fall apart.
TeeDee - I'm afraid I may not have made myself clear. When I say ignore the ridgeline I don't mean that it makes no difference. What I mean is that when you analize the forces you can cut the ridgeline out of the equation. You can look at the forces by looking at the hammock as a black box suspended by two ropes. The only thing that you need to know to calculate the forces on the rope (and rope hardware) is the angle and the load.
Yes, you can replace the hammock by a blackbox, but when you do so, you have to ascribe to the blackbox the properties of the hammock. Assigning zero properties to the ridgeline and saying that the forces are solely determined by the angle of the suspension line is not correct.
SO with or without ridgeline, what matters is the angle of the rope when you are hanging on the hammock.
Incorrect. The ridgeline affects the properties of the hammock and trying to deny that leads to false conclusions.
If you really believe that the ridge line has no affect on the properties of the hammock, then you will have absolutely no hesitation to perform the following experiment:
1. Hang your Hennessy hammock in the way that you like,
2. climb in the hammock,
3. get comfortable,
4. then reach up and with a sharp knife, cut the Hennessy structural ridgeline.
If you are correct, there will zero effect of cutting the ridgeline. If you incorrect, then you will experience the effect of the Hennessy structural ridgeline. Please have your assistant standing by with a video camera and report back the results of your experiment.
txulrich
06-04-2007, 14:54
I can see this as an interesting science experiment. Put force guages on the hammock support ropes. Place 50, 100 and 200 lbs. in a hammock w/o a ridgeline and take the readings. Then try another set with an additional force guage inserted with a ridgeline and repeat the measurements. Anybody got access to a physics lab?????
I don't want to get too technical but to take this argument to conclusion I have to. By reverse engineering my HH I find that the rope has a breaking load of 1600 lbs. The manufacturer's suggested working load on a rope is 20% of that limit (from Samson Ropes literature). That tells me my hammock was designed for a maximum load of 320 lbs on the rope, which at 30 degrees means a 320 lb weight. That results in a safety factor of 1.28 for the design load of 250lbs. If you reduce the angle the factor of safety goes away. You might want to check with the manufacturer what their recommendation is, but as a product developer this is how I make my assessment.
Trying to analyze the forces on the hammock based on the rope strengths is not a very good method for engineering.
hitchman
06-04-2007, 15:22
Please, let's back up TeeDee. I don't think you are understanding me.
Let me follow your example. If I cut the ridgeline the angle of the ropes outside will increase, meaning that the tension on the ropes relaxes. What I stated before is still true. The load on the ropes is a function of the angle. By cutting the ridgeline the angle changed, and so did the tension.
Does this make sense?
Please, let's back up TeeDee. I don't think you are understanding me.
I'm understanding you completely - I don't think the reverse is true. You seem to want to totally deny that the ridge line has any effect. I think you are wrong. I don't think we can agree coming from such totally opposite directions.
We can just leave it at that.
blackbishop351
06-04-2007, 15:49
TeeDee, either you guys aren't understanding each other or he's got you on this one. I've thought about it some more and drew some diagrams. The forces due to the ridgeline - pre-tension and elasticity - can be ignored by simply measuring the sag angle. Basically, what I think hitchman is saying is that a non-ridgeline hammock with a certain sag angle and load will put the same forces on its suspension as a ridgeline hammock with the same load at the same sag angle. Think about it for a minute.
hitchman
06-04-2007, 15:55
Trying to analyze the forces on the hammock based on the rope strengths is not a very good method for engineering.I am interested in the forces that would be applied to the hitchcraft tie when it is used to tension the rope, outside the ridgeline attachment. From that point of view the only thing that I am trying to obtain from this community is the typical range of the rope angle.
I have put many hundreds of hours developing a product that I think has a good application with hammocks. I want to learn as much from this community as possible to be able to make well grounded recommendations.
hitchman
06-04-2007, 15:58
TeeDee, either you guys aren't understanding each other or he's got you on this one. I've thought about it some more and drew some diagrams. The forces due to the ridgeline - pre-tension and elasticity - can be ignored by simply measuring the sag angle. Basically, what I think hitchman is saying is that a non-ridgeline hammock with a certain sag angle and load will put the same forces on its suspension as a ridgeline hammock with the same load at the same sag angle. Think about it for a minute.
Thanks for bringing in some clarity!
Jumping into the end of the arguement.
I think there are not a lot of forces acting on the ridgeline. Here is my thinking (no data to back it up, just my own observations).
I have been playing around with my DIY hammock a lot lately to get the feel just right. A couple times I made the ridgeline so long that it was too long for the length of the hammock. With me in the hammock, the ridgeline would not stay taught. I was at the upper edges of the ridgelines capability.
On another one I used para cord. This stiff has a lot of stretch. I could never get it taught.
When I hang my hammock at a large angle (approx 30 deg), the ridgeline without me in it is taught but not under a lot of preasure. I think that (just by eye-balling) the hammock and hammock support ropes looked to be pretty inline with the suspension straps. If you were to draw a force diagram of this, it would show the hammock pulling down and the support straps pulling up. With an added force from the ridgeline. The bulk of the force would be from the hammock.
I think that the ridgeline would take a lot more force if it was always hung close to 0 deg. I think this would greatly add to the force on it from the hammock. It would also see the force on it from pulling it taught.
I think the best way to test this would be to use some really low rated line as your ridgeline. See what breaks and what doesn't. I think just by looking at the hh ridgeline you can say tom h already knows this. The ridgeline doesn't look strong enough to hold much force.
I think that in the end if he was greatly underestimating the forces added from the ridgeline, his pulleys thingy's would have broken long ago. But since he reported well over 100 uses, I would have to agree with his real word data.
TeeDee, either you guys aren't understanding each other or he's got you on this one. I've thought about it some more and drew some diagrams. The forces due to the ridgeline - pre-tension and elasticity - can be ignored by simply measuring the sag angle. Basically, what I think hitchman is saying is that a non-ridgeline hammock with a certain sag angle and load will put the same forces on its suspension as a ridgeline hammock with the same load at the same sag angle. Think about it for a minute.
Crap beaten again. :mad:
TeeDee, either you guys aren't understanding each other or he's got you on this one. I've thought about it some more and drew some diagrams. The forces due to the ridgeline - pre-tension and elasticity - can be ignored by simply measuring the sag angle. Basically, what I think hitchman is saying is that a non-ridgeline hammock with a certain sag angle and load will put the same forces on its suspension as a ridgeline hammock with the same load at the same sag angle. Think about it for a minute.
Okay - I have thought about this and drawn some diagrams.
Lets look at diagram 1:
http://www.hammockforums.net/gallery/files/2/2/6/ridgeline.03.jpg
This is the classical formula. The force on the suspension is as everybody says. No argument
Now lets add a structural ridgeline:
http://www.hammockforums.net/gallery/files/2/2/6/ridgeline.02.jpg
Now in this case the force F2 is, as before, some force divided by the sine of the angle alpha. The difference is that the force being divided is no longer the weight of the occupant. The force of the ridgeline has entered the scene and altered the angle from beta to alpha. The force f-one in the diagram is now given by the usual formula - no argument. The force f-2 is given by the same formula using the angler beta. But the force used to compute the force F-2 is no longer the weight of the occupant.
The wife is ringing the dinner bell and getting anxious - time to go.
So hitchcraft and BB are right the force on the suspension rope is given by:
D/(2* sin(alpha))
Then I guess where we disagree is that hitchcraft seems to have wanted to say that D is equal to the weight of the occupant and I disagree. Don't have time right now to derive what D is though. The wife is calling for dinner.
Okay - I've been able to think about this over dinner time. What else do you think about??
hitchcraft and BB351 are right. The force D in my previous post is indeed the weight of the occupant . The effect of the ridge line is to amplify the force of the weight of the occupant on the suspension ropes.
But the discussion has proven fruitful for me in another regard. I have been slightly curious about the force on the ridge line, but too lazy and busy to do the analysis. It suddenly occurred to me that the computation of that force is now obvious.
Using the diagram from my previous post:
http://www.hammockforums.net/gallery/files/2/2/6/ridgeline.02.jpg
I can now see that the force on the ridge line is simply:
P = (W/2)(cot(alpha) - cot(beta))
For those mathematically inclined this now gives you a method of estimating the force on your ridge line. If you desire to do so.
Simply lay in the hammock and estimate the two angles alpha and beta or have someone do it for you. Plug your weight and angle values into the formula and you have the force acting on the ridge line.
For my HH Safari, I have estimated alpha at approximately 10 deg. and beta at approximately 60 deg. The ridge line amplifies the force considerably. With my weight at 170 lbs, the force on the ridge line computes out to approximately 432 lbs, give or take the error in my estimation of the angles.
Since I have used 3/32" Yale Crystalyne line for the short ridge line, rated at 1,000 lbs, I am probably over the safety factor that hitchman says to use, but less than 1/2 the rated capacity of the ridge line.
Hennessy has rated the Safari at 350 lbs. Assuming the same angles are obtained, the force on the ridge line comes out at 891 lbs. That is quite a bit of force on that ridge line.
That is for my short ridge line. I will have to swap the ridge line and estimate the angles and compute the force for the long ridge line. Since alpha is about the same, but beta is much smaller, the force on the long ridge line will be smaller. From memory, I would estimate the angle at between 45 deg and 50 deg, which would bring the force down to between 397 lbs and 410 lbs. That is way over half the rating of the BPL guy line cord I used for the long ridge line. Now I feel more comfortable that I decided not to use the Mountain Laural Designs guy line cord rated at 200 lbs for the long ridge line.
blackbishop351
06-04-2007, 19:07
I'm glad we're all on the same page now...I hate arguing, and I hate doubting my own analyses even more :p
I'm glad we're all on the same page now...I hate arguing, and I hate doubting my own analyses even more :p
Ahem - "discussing" please :rolleyes: :D
GrizzlyAdams
06-04-2007, 21:42
Y'all covered a lot of ground in an afternoon.
Understanding that my last physics course was, ahem, 30 years ago, I'm not seeing why force applied to the lines before someone gets into the hammock is not part of the picture here.
Speaking of pictures, consider
http://www.hammockforums.net/gallery/files/4/2/3/forces-3_original.png
A line is suspended between vertical supports, and a total tension P is applied before tying things off. P manefests itself as two opposing horizontal force vectors with equal magnitudes whose sum is P.
A downward directed force with magnitude W is applied at the center of the line.
http://www.hammockforums.net/gallery/files/4/2/3/forces-4_original.png
Because the point mass with weight W isn't moving, there is a force with equivalent magnitude and opposite direction to W, caused by the tension placed on the lines to begin with.
According to received wisdom, the magnitude of the tension on the diagonal is found by considering W as the vertical projection of the diagonal's magnitude. But by symmetry, P/2 is likewise the magnitude of the force on the line due to the original tensioning. Thus it seems as though the stress F1 on the diagonal must be (W/2)/sin(alpha) + (P/2)/cos(alpha).
Given W, P, and the distance between posts, ignoring the effects of the rope material, one ought to be able to compute from a force diagram the distance the weight descends, and hence alpha. I don't remember exactly how that goes (given that forces on the diagaonal have opposing directions) and not enough time to work it out from first principles. Supposing I could remember the first principles.
I had some more scribblings on the ridge line, but that's pointless if I'm wrong in the observations above.
Grizz the befuddled
blackbishop351
06-04-2007, 21:46
You're going about it the right way, but I think missing the detail that TeeDee missed earlier - the sag angle alpha will vary depending on the amount of initial tension (for a fixed load, of course). Thus the whole system can be analyzed by considering the hammock/user/ridgeline as a rigid block, because we account for the initial tension by accounting for the sag angle.
GrizzlyAdams
06-04-2007, 22:03
You're going about it the right way, but I think missing the detail that TeeDee missed earlier - the sag angle alpha will vary depending on the amount of initial tension (for a fixed load, of course).
No, I don't think I'm missing that at all. It comes into play when you think about the force vector that opposes the gravitational one. We know its magnitude---W. We know that that vector must be a projection of the tension vector, for tension is the only opposing force here. The projection will be something like T sin(alpha). Held constant at W, when T grows then alpha must shrink.
I'm not facile enough with the force vectors to pin the tension vector down exactly. But I'm convinced with that detail you can compute alpha from the rest of it.
Thus the whole system can be analyzed by considering the hammock/user/ridgeline as a rigid block, because we account for the initial tension by accounting for the sag angle.
I'm not arguing that you can't treat hammock/user/ridgeline as a rigid block. It is clear to me that you can. There is nothing fundamentally different about the analysis on the suspension lines if we take the point mass W, turn it into a rigid bar with mass W and length r, and hang the suspension cords from the endpoints (under the initial tension of course, an admittedly gnarly detail).
My point is that I don't see that you can ignore the inital tension if the object is to compute the overall tension on the suspension lines. Treating the hammock/user/ridgeline as a rigid block doesn't change that. The tension was applied, and exists outside of that block.
Grizz the disputant
blackbishop351
06-04-2007, 22:51
I really hate not being able to explain things clearly...
If you change the initial tension on the ridgeline, the sag angle alpha changes to compensate. Thus the total tension in the suspension remains constant to some reasonable approximation.
GrizzlyAdams
06-04-2007, 23:25
I really hate not being able to explain things clearly...
If you change the initial tension on the ridgeline, the sag angle alpha changes to compensate. Thus the total tension in the suspension remains constant to some reasonable approximation.
Might be true. Some of y'all believe it's true. Probably is true.
What you and the others are saying is that if you measure alpha, and measure W, then alpha is smaller than it would really be if the only force involved here is W. Agreed. Then you say that everything we need to know about tension forces are coded by alpha's value, so much so that we don't need really to figure them at all. That's the "then a miracle occurs" step whose validity I'm not sure of. At one level I can imagine it being true; the vertical force (happens to be gravity) is a geometric projection of the force on the suspension rope (happens to be an aggregate force), knowing the projection and the angle, you can compute the aggregate force.
I'm a first principles kind of guy, and if it's true then it should fall out of the math. I tinkered around with forces and such, I saw that exactly the same argument for computing what the stress on the suspension rope is due to weight ( (W/2)/sin(alpha)) has a symmetric argument when you think not about the vertical gravitational force, but about the horizontal tension force. Seemed there'd be a way to compute the actual force on the suspension rope, by decomposing it into its constituent parts.
So I figure that if I can dot all the i's and cross all the t's and nail the relationships between the tension force, the weight, and the angle, then I can understand why this trick works, if indeed it does.
Grizz the pedantic
blackbishop351
06-04-2007, 23:42
I'd love to continue the discussion, but maybe we shouldn't clutter the open Forums with it...PM me if you want?
angrysparrow
06-05-2007, 07:29
I'd love to continue the discussion, but maybe we shouldn't clutter the open Forums with it...PM me if you want?
Please don't do that. This is clearly a hammock-related topic, and I quite enjoy following along.
NCPatrick
06-05-2007, 07:42
Agreed. Even though I have no idea what you guys are talking about, please proceed.
blackbishop351
06-05-2007, 08:37
OK OK OK...
Let x be the length of the suspension from the ridgeline attachment point to the tree. Let T be the initial tension applied to the system (along the ridgeline/suspension).
Since the materials are elastic (ALL materials are, to some approximation), we can model their static properties using the harmonic oscillator equation F = -kx . This is kind of back-of-the-napkin (as I'm sure an engineer would tell you) but should suffice for our macro-scale purposes here. Here, F is the restoring force due to displacement, so whatever tension T applied to the system must oppose this in order to stretch the materials. Also, the spring constant k is an inherent property of the material, which would have to be determined experimentally to achieve a thorough analysis. Fortunately, we don't need to do that for a broad-strokes analysis like this.
In order for the system to be static after the initial tension T is applied, we must then have T = kx for some stretch amount x . This then gives x = T/k for a given tension.
Upon adding a load to the hammock, we know that the tension in the materials will change by an amount delT = W/sin(alpha), where W is the user's weight and alpha is the sag angle yet to be determined. Combining this with the previous result, we have T + delT = k(x + delx), where delx is the resulting change in length of the suspension.
Because of the geometry involved - the stretching changes the length, which allows a right triangle with apex angle alpha - we also have cos(alpha) = x/(x + delx), which can be substituted to give Tsin(alpha) = Ttan(alpha) - W. This is transcendental in alpha, but still gives a functionality between sag angle alpha, user weight W, and initial tension T.
What this boils down to is something that's common sense - if you add a weight to a tensioned line, the line will sag. Increase the weight (holding the tension constant) and the line sags more. Increase the tension (with the weight constant) and the line sags less. We can see now that the sag is actually due to a change in length of the (necessarily) elastic material.
I hope that helps.
It looks like Jeff and Troll need to add greek letters to this site. The lack of them is really hurting the look of your equations. Us engineers don't have those issues you physics book readers do.;)
blackbishop351
06-05-2007, 10:22
I know, right? We need TeX compatibility... :p
GrizzlyAdams
06-05-2007, 11:49
OK OK OK...
...
In order for the system to be static after the initial tension T is applied, we must then have T = kx for some stretch amount x . This then gives x = T/k for a given tension.
I think for consistency with what you say later, really this is
T = k*delx
Upon adding a load to the hammock, we know that the tension in the materials will change by an amount delT = W/sin(alpha), where W is the user's weight and alpha is the sag angle yet to be determined. Combining this with the previous result, we have T + delT = k(x + delx), where delx is the resulting change in length of the suspension.
Because of the geometry involved - the stretching changes the length, which allows a right triangle with apex angle alpha - we also have cos(alpha) = x/(x + delx), which can be substituted to give Tsin(alpha) = Ttan(alpha) - W. This is transcendental in alpha, but still gives a functionality between sag angle alpha, user weight W, and initial tension T.
Let's go slow here. x is the length before any kind of tension. Putting tension on the
line stretches it by a length d, which gets wrapped around the tree or pulled past the
grip point of my rings or your buckles. I'm going to approximate this by assuming instead that
I dig up the tree and plant it a distance d further away, making a length x+d to reflect the
original tension T. Since T = k*d, we can use d = T/k.
Now I hang the hammock and see an angle alpha. The length of the line
from tree to hammock is
( x+(T/k) )/cos(alpha)
This means that the line has stretched ( x+(T/k) )/cos(alpha) - x )
so that the restoring force (complete tension) vector is along the
line, pointed at the tree, with magnitude
k*((x+(T/k))/cos(alpha) - x )
We know that the horizontal projection of this force exactly counter-balances the
downward force W due to hanging the hammock/user. That gives us
W = sin(alpha) * k * ((x+(T/k))/cos(alpha) - x )
So in principle we can express alpha in terms of W,T, k. But, interestingly,
also the original length x is showing up. We've not seen that before in our
back-of-the-napkin analyses.
But then a miracle does indeed occur . We re-arrange the equation above to
W/sin(alpha) = k * ((x+(T/k))/cos(alpha) - x )
The left-hand-side is the equation you guys like. The right-hand-side
is the magnitude of the tension force on the suspension rope, which is what I was wanting to compute.
I am contented now, thanks for the patient explanations.
Grizz
I know, right? We need TeX compatibility... :p
Your just jealous I write my whole thesis without a single equation in it.:p :D
sk8rs_dad
06-05-2007, 12:03
Any idea where can I order some massless rope and maybe a frictionless pulley too? ;)
NCPatrick
06-05-2007, 12:18
I'm going to approximate this by assuming instead that
I dig up the tree and plant it a distance d further away
Now, now... if everyone solved theoretical problems like this, there would be no theoretical trees left. ;)
GrizzlyAdams
06-05-2007, 12:25
In retrospect, we don't need to dig that tree up.
One thing that makes the analysis go is that the tension is a vector pointed along the suspension line with magnitude T, which is what the engineering guys knew intuitively but I had to be told. The other thing that makes it go is that the counter-force to gravity on the hammock/user is sin(alpha)*T, so W = sin(alpha)*T, no matter what T is. So we can leave the tree alone, T will be a little bit different, but the same principle applies.
leaving no trace,
Grizz
hacktorious
06-05-2007, 13:16
Does Hook's Law really apply here? Wouldn't it be more accurate to break it down into x and y components and solve it like a simple truss, or cable problem?
Hmmm, maybe I am just rambling again. It has been a while since I have had to solve a problem like this.
blackbishop351
06-05-2007, 13:23
In retrospect, we don't need to dig that tree up.
One thing that makes the analysis go is that the tension is a vector pointed along the suspension line with magnitude T, which is what the engineering guys knew intuitively but I had to be told. The other thing that makes it go is that the counter-force to gravity on the hammock/user is sin(alpha)*T, so W = sin(alpha)*T, no matter what T is. So we can leave the tree alone, T will be a little bit different, but the same principle applies.
leaving no trace,
Grizz
I wouldn't phrase it quite that way - gravity induces tension, not the other way around - but yeah you're right. I need to stop doing math early in the morning.
I am trying to obtain from this community is the typical range of the rope angle.
As tight as I can get it. I am short so the tighter I make it the less I have to stand on my tippies. Also I got a stand last night (yay me) and I have to pull tight as I can to stay off the floor. My wife let me set it up and sleep in the family room last night.
txulrich
06-05-2007, 15:37
Any idea where can I order some massless rope and maybe a frictionless pulley too? ;)
I had a set, but I loaned to Jeff last month. You might go and ask him for them.
..... From that point of view the only thing that I am trying to obtain from this community is the typical range of the rope angle.
....
hitchcraft - browse through this thread (http://www.hammockforums.net/forum/showthread.php?t=711). It discusses how tight people pull their HH when hanging - it's been awhile since I last read it, but if I remember correctly, the consensus seemed to be that most (all?) pulled their suspension as tight as they reasonably could.
At least 2 reasons for that: get bug netting out of the face, and if you don't pull tight then you have to put the attachment points to the trees high and the greater the tree separation, the higher the attachment points. I have written about that second point in the first post on this thread (http://www.hammockforums.net/forum/showthread.php?t=1206). It quickly gets to the point that you can no longer reach the attachment point if you don't pull the suspension line tight when hanging or you have your butt bouncing on the ground all night.
hitchman
06-06-2007, 20:24
It seems that a number of users like to tighten the hammock as much as possible. It's still hard to put a value on it. When several of the elements connected are somewhat flexible a hammock that is horizontal by itself with sag. What the angle of the rope ends up being is still unknown. Most of the pictures I see lead me to believe that 30 degrees is typical, but I understand not everybody will agree to that. In any case a tighter angle will start overloading the hammock components at some point.
for a 200lb weight we have the following tension versus the angle:
30 degrees - 200 lbs
20 degrees - 292 lbs
15 degrees - 386 lbs
10 degrees - 575 lbs
5 degrees - 1147 lbs
If the hammock starts at shoulder height and the ropes are 3 feet long (to the ends of the ridgeline) then at 30 degrees the ridgeline will sag about 1.5 feet, and it will be about 4 feet from the ground. I think that is enough to keep the hammock from touching down. If the trees are further than 13-14 ft apart then the attachment height needs to be higher or the angle smaller.
Okay, maybe this is too much theorising, but after the mathfest that went on earlier in my absense I just couldn't help keepint the ball rolling.
for a 200lb weight we have the following tension versus the angle:
30 degrees - 200 lbs
20 degrees - 292 lbs
15 degrees - 386 lbs
10 degrees - 575 lbs
5 degrees - 1147 lbs
This may be a silly question but remember I am not a math or physics guy.
If the angle is greater than 30 degrees (45 thru 90) does the tension level out and remain to equal the weight being applied?
blackbishop351
06-06-2007, 21:23
This may be a silly question but remember I am not a math or physics guy.
If the angle is greater than 30 degrees (45 thru 90) does the tension level out and remain to equal the weight being applied?
A 90* hang is impossible, but yes the tension would be (actually twice) the weight of the user if it could be done.
A 90* hang is impossible, but yes the tension would be (actually twice) the weight of the user if it could be done.
Yeah...... 90 was a bad example (this is what happens when I venture into the math world).
But at 45 degrees how much tension would 200lbs apply?
blackbishop351
06-06-2007, 22:36
Actually i misspoke myself...at 90*, the TOTAL tension would be the user's weight. At 30*, it's twice the user's weight. And 45* would be somewhere in between, of course. By total tension, I mean we account for both ends of the hammock. The numbers hitchman listed above only count half the system.
GrizzlyAdams
06-06-2007, 23:38
This may be a silly question but remember I am not a math or physics guy.
If the angle is greater than 30 degrees (45 thru 90) does the tension level out and remain to equal the weight being applied?
The formula Hitchman is using computes the tension on one of the suspension ropes (which makes sense when you're interested in the threat of one of these ropes breaking.) After much weeping and gnashing of teeth it was established that that formula is
tension = (W/2)/sin(angle)
where W is the weight. You can punch this up on any calculator. With W = 200 lbs and angle = 45 degrees you get tension = 141.4 lbs.
Grizz the teary-eyed
I think this one is a good illustration:
http://whiteblaze.net/forum/vbg/showimage.php?i=13792&catid=member&imageuser=1908
Thanks, Smee!
GrizzlyAdams
06-07-2007, 06:07
I think this one is a good illustration:
http://whiteblaze.net/forum/vbg/showimage.php?i=13792&catid=member&imageuser=1908
Thanks, Smee!
The whiteblaze thread where Smee posted this is a hoot.
http://www.whiteblaze.net/forum/showthread.php?t=18685
Starts with similar questions about stresses, and spins wildly out of control, due in no small part to some of the posters in the hammockforums.
The mind staggers at the possibilities.
Grizz, restraining self
Actually i misspoke myself...at 90*, the TOTAL tension would be the user's weight. At 30*, it's twice the user's weight. And 45* would be somewhere in between, of course. By total tension, I mean we account for both ends of the hammock. The numbers hitchman listed above only count half the system.
Ahhhhh ok. I did not realize he was only accounting for half the system. That is where I got confused. It looked like the tension would be less than the initial weight after 30*.
GrizzlyAdams
06-07-2007, 09:28
Dear Hammock Physics 102 class,
A number of you have come to my office hours asking about Professor TeeDee's cryptic solution to midterm question number 42. Recall that the problem asks for you to give the tension on a ridgeline, given inclination angle alpha on the suspension ropes above the ridgeline, inclination angle beta on the suspension ropes below the ridgeline, and weight w. Professor TeeDee's solution is illustrated below
http://www.hammockforums.net/gallery/files/4/2/3/hammock-physics-1.png
The key to understanding this solution is to think about the forces on the join point J where the ridgeline attaches to the suspension ropes.
http://www.hammockforums.net/gallery/files/4/2/3/hammock-physics-2_251656.png
Slightly abusing notation by reversing direction, the force F2 is the restorative force due to stretching, the force F1 is a force due to gravity on the hammock/user along the rope, and force P is the tension on the ridgeline. We know how to compute F2; we know that the force due to gravity is equal in magnitude to the restorative force on that line. Importantly, we know that in vector arithmetic P + F2 + F1 = 0; J is stationary, so the forces cancel.
Remember the geometry of vector addition : to compute F2+F1 we can draw vector F2, reposition a vector with the same length and angle as F1 at the head of F2, and the sum is the vector extending from F2's base to the head of the repositioned F1. This is illustrated below.
http://www.hammockforums.net/gallery/files/4/2/3/hammock-physics-3.png
The first thing to note is that the direction of the vector F2+F1 has to be opposite to the direction of P. This is handy, because it means we can use right triangles in the analysis.
Now you need to recall some geometric identities, also illustrated in this picture, of angles equal to alpha and beta. You also need to recall a little bit of trigonometry, that says the length of the adjacent side is the cosine of the angle times the length of the hypotenuse. Knowing the length F2 and the angle alpha, we can compute the length of line segment ac; knowling length F1 and the angle beta, we can compute the length of line segment bc. The difference between lengths of ac and bc gives the length of ab, which is the magnitude of P, and is equal to Professor's TeeDee's solution.
Professor TeeDee reminds you all that the final exam will ask this same question, without giving you angles alpha and beta, only the spring constants and lengths of the suspension ropes and ridgeline.
Please note that June 15 is the last day you can drop this class without penalty. There are still plenty of spaces available in Professor JustJeff's popular survey course on Hammock Camping.
Hammock Physics 102 TA
NCPatrick
06-07-2007, 09:48
Please note that June 15 is the last day you can drop this class without penalty. There are still plenty of spaces available in Professor JustJeff's popular survey course on Hammock Camping.
Hammock Physics 101 TA
May I continue to audit your class? My plate is kinda full right now with the 'Whittling, Rocking, Whistling and Spitting' class I just signed up for. It is independent study, but it may take up a lot of my time. :D
May I continue to audit your class? My plate is kinda full right now with the 'Whittling, Rocking, Whistling and Spitting' class I just signed up for. It is independent study, but it may take up a lot of my time. :D
I am planning on signing up, that way if I ace the tests I'll stay in. I just need to find out the last day to drop without taking an incomplete.:eek:
Bug-Bait
06-07-2007, 10:52
Please note that June 15 is the last day you can drop this class without penalty. There are still plenty of spaces available in Professor JustJeff's popular survey course on Hammock Camping.
Hammock Physics 102 TA[/QUOTE]
**** you Grizzly...and I thought that I would finally stop having those dreams about having forgotten to go to class all semester and didn't study for the exam. Problem is, I've been out of college for 35 or so years and still occasionally have that dream...lol
Michael
qpens
hitchman
06-07-2007, 11:53
This may be a silly question but remember I am not a math or physics guy.
If the angle is greater than 30 degrees (45 thru 90) does the tension level out and remain to equal the weight being applied?
Dgrav, here's my delayed response - you probably got the answer in the meantime.
At 45deg , a 200lb load would produce 141 lb tension on each rope. If the hammock is level the load is equally split by the two ends, which is my assumption. Being a symmetric system I just have to look at one half, meaning that half the vertical load is sustained by each end. I know this because if I tie only one end the hammock won't stay horizontal:D .
hitchman
06-07-2007, 12:01
Dear Hammock Physics 102 class,
... There are still plenty of spaces available in Professor JustJeff's popular survey course on Hammock Camping.
Hammock Physics 102 TA
Pretty impressive stuff. Being a lazy engineer I try to simplify things first until I get a very simple formula like: w*2*0.5=w
Of course, you might say that a black box may not be as comfortable as a hammock with a ridgeline. That's why engineers solve small problems one at a time while physicists are still trying to solve one big problem: how does the universe work?
Pretty impressive stuff. Being a lazy engineer I try to simplify things first until I get a very simple formula like: w*2*0.5=w
Of course, you might say that a black box may not be as comfortable as a hammock with a ridgeline. That's why engineers solve small problems one at a time while physicists are still trying to solve one big problem: how does the universe work?
The engineer in me does something like that too. I figure out the best I can easily what I need. Then throw a huge saftey factor on to it. That usually gets me to where I want to be. Then again, all my grad work is with bridges where everything is way over designed. With how light the spectra cord is now, I really don't think I am adding any extra weight.
Dear Hammock Physics 102 class,
A number of you have come to my office hours asking about Professor TeeDee's cryptic solution to midterm question number 42. Recall that the problem asks for you to give the tension on a ridgeline, given inclination angle alpha on the suspension ropes above the ridgeline, inclination angle beta on the suspension ropes below the ridgeline, and weight w. Professor TeeDee's solution is illustrated below
http://www.hammockforums.net/gallery/files/4/2/3/hammock-physics-1.png
The key to understanding this solution is to think about the forces on the join point J where the ridgeline attaches to the suspension ropes.
http://www.hammockforums.net/gallery/files/4/2/3/hammock-physics-2_251656.png
Slightly abusing notation by reversing direction, the force F2 is the restorative force due to stretching, the force F1 is a force due to gravity on the hammock/user along the rope, and force P is the tension on the ridgeline. We know how to compute F2; we know that the force due to gravity is equal in magnitude to the restorative force on that line. Importantly, we know that in vector arithmetic P + F2 + F1 = 0; J is stationary, so the forces cancel.
Remember the geometry of vector addition : to compute F2+F1 we can draw vector F2, reposition a vector with the same length and angle as F1 at the head of F2, and the sum is the vector extending from F2's base to the head of the repositioned F1. This is illustrated below.
http://www.hammockforums.net/gallery/files/4/2/3/hammock-physics-3.png
The first thing to note is that the direction of the vector F2+F1 has to be opposite to the direction of P. This is handy, because it means we can use right triangles in the analysis.
Now you need to recall some geometric identities, also illustrated in this picture, of angles equal to alpha and beta. You also need to recall a little bit of trigonometry, that says the length of the adjacent side is the cosine of the angle times the length of the hypotenuse. Knowing the length F2 and the angle alpha, we can compute the length of line segment ac; knowling length F1 and the angle beta, we can compute the length of line segment bc. The difference between lengths of ac and bc gives the length of ab, which is the magnitude of P, and is equal to Professor's TeeDee's solution.
Professor TeeDee reminds you all that the final exam will ask this same question, without giving you angles alpha and beta, only the spring constants and lengths of the suspension ropes and ridgeline.
Please note that June 15 is the last day you can drop this class without penalty. There are still plenty of spaces available in Professor JustJeff's popular survey course on Hammock Camping.
Hammock Physics 102 TA
Gee - you make it seem so easy - all I did was subtract the horizontal forces.:D
Your solution is better for the Freshman textbook though. :rolleyes:
GrizzlyAdams
06-07-2007, 15:23
Gee - you make it seem so easy - all I did was subtract the horizontal forces.:D
and if you'd said that in your post, I wouldn't have understood what that implied. Still don't. Had to work it out from the few principles I know.
Your solution is better for the Freshman textbook though. :rolleyes:
that's appropriate, seeing as my physics coursework was limited to the Introductory course. I remember spending a whole afternoon trying to work out why the equation for the differential cross-section in an Rutherford scattering experiment was correct. The smart guys just accepted the equation, knew when to use it, and got on with things.
it's been fun,
Grizz
and if you'd said that in your post, I wouldn't have understood what that implied. Still don't. Had to work it out from the few principles I know.
ahhhh -
bc = horizontal force due to person sitting in hammock
ac = horizontal force due to person sitting in hammock + horizontal force due to tension in structural ridge line
ac - bc = horizontal force due to tension in structural ridge line
I was looking at and visualizing the forces acting and that's what I saw while thinking about it at dinner. Then had to sit down and remember my trig and figure what ac and bc were and there I was. Looking at the forces acting is what finally convinced me that hitchcraft and BB351 were right - only one vertical force acting - occupant weight + hammock weight.
GrizzlyAdams
06-07-2007, 17:40
ahhhh -
bc = horizontal force due to person sitting in hammock
ac = horizontal force due to person sitting in hammock + horizontal force due to tension in structural ridge line
ac - bc = horizontal force due to tension in structural ridge line
I was looking at and visualizing the forces acting and that's what I saw while thinking about it at dinner. Then had to sit down and remember my trig and figure what ac and bc were and there I was. Looking at the forces acting is what finally convinced me that hitchcraft and BB351 were right - only one vertical force acting - occupant weight + hammock weight.
Shiny!
At our house dinner time conversation is usually steered by my better half towards literature, religion, politics, and current events.
(while I'm thinking about whatever technical problem I happen to be working on that day. ;) )
Grizz
I finally measured the angles on my hammock with the two ridge lines. My "eyeball" estimates were off by a factor of at least 2.
angle beta for the long ridge line: 29.8 deg (less sag)
angle beta for the short ridge line: 38.3 deg (lots and lots of sag)
angle alpha for both: 20.6 deg
That gives me the following tension force on the ridge line:
long ridgeline: 78 lbf - 156 lbf total (cord rated at 563 lbf)
short ridge line: 119 lbf - 238 lbf total (cord rated at 1,000 lbf)
The ridge line cords I've used are way over what is necessary. :D That puts me well within any safety factor the engineers would want.
I could have used the Mountain Laurel Designs guy line cord for the long ridgeline and the BPL cord for the short ridgeline.
I ordered the Mini RopeTie's on the evening of 6/6 and got them on 6/9. Very fast service.
Set up on my Expedition Hennessy was real quick using the instructions. The 1/4 inch spectra is at the top of the Mini's range so it was tight going in but that also keeps it from going down into the grooves very far, which should be beneficial to the spectra over time.
Hitchcraft said he was comfortable using the Mini's with his 180# so I stuck my 180# in the hammock. So far, so good. Once I laid there for a bit and got my confidence up I started flopping around a bit, then harder, then even harder and they were good to go.
Normally when I lash the hammock I have to sit on it to take the slop out and then tighten it again. Then as the night passes the sag still increases. With the HitchCraft, I set it and it stayed right there all night.
After years of lashing these things are great. Fast, easy, light and cheap.
Final analysis: I'll have to keep an eye on them over time and see how they do in the long haul and do some winter tests. But for right now I believe these little jewels come in somewhere just below pizza and beer.
hitchman
06-11-2007, 16:26
...
Normally when I lash the hammock I have to sit on it to take the slop out and then tighten it again. Then as the night passes the sag still increases. With the HitchCraft, I set it and it stayed right there all night.
After years of lashing these things are great. Fast, easy, light and cheap.
Final analysis: I'll have to keep an eye on them over time and see how they do in the long haul and do some winter tests. But for right now I believe these little jewels come in somewhere just below pizza and beer.
I take it that's an endorsement. Thanks for trying my product. The key is not to overtension the hammock (under 30 deg loaded) since it will drive the load beyond the safe working load. I would be interested in getting wear data over time.
Just Jeff
06-11-2007, 17:11
Holy crap, you guys are smart.
Mathematicians look for "truth"...this time I'll settle for "true enough that I don't fall on my butt." In the search for "true enough," as you might have heard in the other threads I haven't read yet, I handed out a few pair of these Hitchcrafts at the Hot Springs SEHHA this weekend, courtesy of Hitchcraft. The testers will report their conclusions back to this site...when they're ready I'll consolidate them so it's easy to find.
Thanks, Hitchcraft!
hitchman, if you come up with a product that will do straps instead of rope I would be interested in a set.
I take it that's an endorsement. Thanks for trying my product. The key is not to overtension the hammock (under 30 deg loaded) since it will drive the load beyond the safe working load. I would be interested in getting wear data over time.
Definately an endorsement. I bounced in the hammock until I was affraid I'd rip it. So I'm pretty sure they'll stand up to my normal hanging. I normally don't get off the bike till dusk*, so these are going to make setting up in the dark a lot faster and easier.
(* it's harder for the land owners to spot me then. :D )
hitchman
06-11-2007, 20:25
hitchman, if you come up with a product that will do straps instead of rope I would be interested in a set.
Thanks for the vote of confidence. However that is not likely to happen anytime soon. As it is I jumped off the deep end with this design which is my first attempt to sell my own product. I started with a lot of optimism but the reality is that it takes a lot of marketing resources to get the exposure and sales to break even. At this point I don't think about it and try to be satisfied that it is a learning experience, and as you know going to school is expensive.:) All the retailers I contact are carrying the Nite-Ize figure-9, which is not nearly as functional (IMMO), and Made in China trumps the better product.
Well, sour grapes...:rolleyes:
I ordered the minis and they arrived within 6 days. Not bad for overseas. I've done the same as Btourer and gingerly tested my weight then bounced. They work exceptionally well, very easy to setup and adjust on my Speer hammock. I'll be getting more for the other hammocks I'm making.
Definately an endorsement. I bounced in the hammock until I was affraid I'd rip it. So I'm pretty sure they'll stand up to my normal hanging. I normally don't get off the bike till dusk*, so these are going to make setting up in the dark a lot faster and easier.
(* it's harder for the land owners to spot me then. :D )
bbtourer - do you pull the suspension tight when you hang or do you hang loose with a lot of sag in the suspension - even before getting in the hammock - as pictured by hitchman?
Cannibal
06-12-2007, 14:54
After spending the last 3 hours reading this thread; I think I just heard something go "pop" in my brain! I hope it wasn't anything important.:confused:
After spending the last 3 hours reading this thread; I think I just heard something go "pop" in my brain! I hope it wasn't anything important.:confused:
Nah - it's like popcorn, just overheated.:D :rolleyes:
hitchman
06-12-2007, 19:22
I ordered the minis and they arrived within 6 days. Not bad for overseas. I've done the same as Btourer and gingerly tested my weight then bounced. They work exceptionally well, very easy to setup and adjust on my Speer hammock. I'll be getting more for the other hammocks I'm making.
I'm glad you received them already. It's the first order I had from AUS.
I hope the parts work for you. I assume you are not using 1/8" line, for which I do not recommend the Mini ropetie. The margin of safety for the 1/8 line is significantly lower than with 3/16 or 1/4, and it can slip at the jam cleat, when overloaded. One way to ad safety margin is to use two ropeties in tandem as shown in the user guide card... that will also ad leverage to the tensioning, but working so close to the limit will reduce the wear life of the parts.
Hitchman, I use a Speer type setup with 3/16 ropes. Today I put an electronic strain guage on one of my ropes and got max 85kg (approx 187lb). I feel comfy with that so she'll do me.
bbtourer - do you pull the suspension tight when you hang or do you hang loose with a lot of sag in the suspension - even before getting in the hammock - as pictured by hitchman?
I pull the hammock until it's pretty snug but not overly so. Generally the farther apart the trees are the tighter I pull it. I'd actually give a bit more than I'd like to compensate for the lashing giving. For this test I had it hung between trees that were about 11 feet apart and I had the ridgline pulled tight enough to twang when you plucked it. I intentionally hung between closer trees so that I'd be getting less sag. I'd say I was at about 25 degrees angle.
I got mine from Cabelo's in the mail yesterday. After seeing them I am kind of skeptical of using them now. I am not to thrilled about the pinching points that pinches the rope to keep it from slipping.
hitchman
06-21-2007, 15:34
I got mine from Cabelo's in the mail yesterday. After seeing them I am kind of skeptical of using them now. I am not to thrilled about the pinching points that pinches the rope to keep it from slipping.
To clarify, atroll is taling about the zig-zag gizmo not the Hitchcraft tie.
New to the forums and I'm intrigued by this device.
Any more long term testing going on?
Who's using this? Seems simple enough.
-bmike
We are still using them for my kids as they are they are a lot lighter than myself. I've gone to the large Fig 9's as they pack flat. To date the kids find it easy to use the hitchcraft - I'm talking about the small ones. Just got back from 4 days hike with no problems at all with the HC's, also no probs with damage to the lines we are using, about 4 or 5 mm.
warbonnetguy
10-15-2007, 19:02
Hitchman, I use a Speer type setup with 3/16 ropes. Today I put an electronic strain guage on one of my ropes and got max 85kg (approx 187lb). I feel comfy with that so she'll do me.
good job kiwidad, i was wondering if someone would actually check all that math with a real world experiment. never heard of an electric strain tester before, sounds pretty cool, and expensive.
did the max force include getting in and bouncing around?did you bounce around at all? how much do you weigh? does the 187 seem to be about right with what hitchman predicted for the weight and sag angle and resulting force on the suspension rope? sounds like maybe those guys did figure out the right equation to use. that is alot less than the amount of force i was always was told was on each rope.
The setup was Speer with no ridgeline. When I bounced it didn't add that much extra. The strain guage was one I borrowed from work and very accurate. I put it in line with the hitchcraft. Probably the only thing I am careful about is when I pack the hammock away into a bag that the HC doesn't stab into the pack as the ends are pointed. As for calculations vs actual, I only tested for my setup. I can't vouch for other setups as I don't have access to other setups.
warbonnetguy
10-15-2007, 19:38
i only meant compared to the figures those math guys came up with earlier in the thread. i couldn't tell, b/c you didn't mention your actual weight or the sag angle you were hanging from.
basically it said the amount of force on each rope is equal to the weight of the user when the suspension rope angle is 30 degrees,
so i guess i'm just curious if you were hanging more than, less than, or about 30 deg? and how much you weigh.
I wish I could borrow strain guages from work. I'd be sooooo thrilled.
I love that video.
The device reminds me of my current system:
http://l2i.org/hammock_tensioning%20block.jpg
http://l2i.org/hammock_tensioning%20block.jpg
But yours looks to work better for only 5g heavier.
I may just buy a pair! Do you ship to Hong Kong?
i only meant compared to the figures those math guys came up with earlier in the thread. i couldn't tell, b/c you didn't mention your actual weight or the sag angle you were hanging from.
basically it said the amount of force on each rope is equal to the weight of the user when the suspension rope angle is 30 degrees,
so i guess i'm just curious if you were hanging more than, less than, or about 30 deg? and how much you weigh.
I weigh about 80kg and got a reading of 85kg. Angle about 30-40deg. Unfortunately I can't try it again as one of the guys at work have busted the gauge, a pity as it is an expensive piece of equipment.
Bummer...
But your single data point certainly indicates the right range.... Slightly more than your bodyweight in tension on each end. :)
mataharihiker
10-27-2007, 16:40
There is nothing like a thread with math, engineering and gadgets to get me all warm and tingly...I just finished reading all 12 pages, believe it or not...I love people who know what they're talking about...I have bad hands, as soon as I get my Clark NA and see the rope diameter, I'm ordering one of hitchman's pulley's...after all, you only need to use one so the weight is insignificant...I weigh 110 so I couldn't break one if I tried....
Talking about that stuff get me warm and tingly... But designing things makes me REALLY happy. :D
Strapman
12-14-2007, 13:55
Since we are talking math check out this equation:
a = b
ab = b2
ab a2 = b2 a2
a(a b) = (a + b)(a b) [(a - b) on both sides cancel each other out]
a = a + b [remember that a = b]
a = a + a
a = 2a
1= 2
GrizzlyAdams
12-14-2007, 14:11
Since we are talking math check out this equation:
a = b
ab = b2
ab a2 = b2 a2
a(a b) = (a + b)(a b) [(a - b) on both sides cancel each other out]
a = a + b [remember that a = b]
a = a + a
a = 2a
1= 2
oh my. If Sister Mary Augustine catches you dividing by zero she'll whack your knuckles with a ruler. Hurts, 'cause its made of cast iron!
Grizz
I don't see how you got from ab – a² to a(a – b) since subtraction isn't commutative [not same as a(b – a)].
GrizzlyAdams
12-14-2007, 14:14
I don't see how you got from ab aČ to a(a b) since subtraction isn't commutative [not same as a(b a)].
multiplied both sides of the equation by -1
then risked incurring SMA's wrath and wruler.
Grizz
NCPatrick
12-14-2007, 14:21
Sister Mary Augustine? That's not wrath, she just loves inflicting pain... pain is weakness leaving the body.
GrizzlyAdams
12-14-2007, 15:23
Sister Mary Augustine? That's not wrath, she just loves inflicting pain... pain is weakness leaving the body.
SMA : Fictional 9th grade parochial school math teacher known to haunt the bridge hammock thread, correcting the math errors frequently found there.
Grizz
NCPatrick
12-14-2007, 15:26
Sorry, I must have been thinking of 'the Penguin', Sister Mary Stigmata, of Blues Brothers fame. Let's get back to the math.
angrysparrow
12-14-2007, 15:31
Sorry, I must have been thinking of 'the Penguin', Sister Mary Stigmata, of Blues Brothers fame. Let's get back to the math.
Nobody should ever apologize for invoking a Blues Brothers reference! :D
Strapman
12-14-2007, 15:44
My Father (the retired highschool physics and chemistry teacher) used to love driving kids crazy with that one. I think he stopped when one student became convinced that "math is farce" and alomst dropped out of school. :confused:
as far as sisters go, i favor sister mary elephant (cheech and chong) ?? no one remembers her?
pure_mahem
12-14-2007, 18:37
Is she the one getting the cavity search at the border,LOL!
GrizzlyAdams
12-14-2007, 20:39
as far as sisters go, i favor sister mary elephant (cheech and chong) ?? no one remembers her?
class. Class. Claaaassss. SHUT UP! Tank you.
don't remember much from those days, but the C&C skits are burned into the memory.
Grizzled Grizz.
BillyBob58
12-14-2007, 22:00
class. Class. Claaaassss. SHUT UP! Tank you.
don't remember much from those days, but the C&C skits are burned into the memory.
Grizzled Grizz.
Without question! Blind Melon Chitlin! "Goin down town, gonna see my gal, gonna sing her a song..................". Yall should know the rest!
you got it griz, "class...give me that knife! shhssssss...thonkkkk, I meant hand it to me"
Blind Melon was good BillyBob, but my fav was the dog skit:
"hey man, lets go look for Fifi, I think that poodles in heat"
" no man, lets go out to the highway and chase cars, get high on the exhaust fumes"
"aw man, that sh#ts bad for your health"
"you mean the exhaust fumes are bad for you"
"no man, its bad for you when the wheel runs over your head"
Sorry, got off track of the thread...what can I say - flashbacks
slowhike
12-15-2007, 07:40
OK class... back to hammocks.
LOL, Grizz - you are in the zone man!
Thanks to this forum, I have found numerous ideas and we I came upon this thread I ordered the Hitchcrafts (the small version). Tried them out this past weekend with my HH and they worked great! I really like them over the lashing technique.
hacktorious
01-30-2008, 17:44
Well, I gave in and got a couple mini-hictchcraft's as well. I just tried them out, and I have to say, they are AWSOME!!!!!
FishBone
11-21-2008, 13:07
I handed out a few pair of these Hitchcrafts at the Hot Springs SEHHA this weekend, courtesy of Hitchcraft. The testers will report their conclusions back to this site...when they're ready I'll consolidate them so it's easy to find.
Thanks, Hitchcraft!
I'm considering using the mini's on my HH ULBA. Bringing this up again to see what the testers said. (I couldn't find the consolidated results.)
How'd they do?
I forgot about the mini hitchcrafts - I just ordered 4 of them - 2 for my blackbird, and 2 for my bridge. I love the monster rope ties for my Clark, so I'm going to give these a go. A little light at 250 lbf, but I'll give it a shot just because they are so easy to use and adjust. Some of the rings I've used in the past had lighter load ratings than that. :eek:
hacktorious
11-21-2008, 17:47
I have been using them for a while. They are awesome. I have not found a system that compares.
6 feet over
11-21-2008, 23:41
I hate math.:D
...want my hammock ridiculously over-engineered, though. My backside, and my ego, don't like taking falls. ;)
I too hate math, yet my 12 year old son is two grades advanced and shows no sign of slowing down. (No, he doesn't resemble the mailman or the UPS guy!:mellow:)
Add me to the ranks of those that want his hammock ridiculously over-engineered. I'm perfectly willing to carry a little more weight to assure my peace of mind.
6
canalvoyageur
11-22-2008, 05:04
Hi there guys and gals, First I am to some of you a relative beginner, have read the post on tensioning, let me relate how I tension my Hammock, Incidentally I not apposed to some of the Hitches, or Knots employed.
My Hammock is what is termed as gathered end to which I cut off the long straps, first forming a simple loop to which I attached a carribener, from the excess strap I tied a tape knot and used one to go around the tree this being the foot of me Hammock, now to the other tree I use a Canoe strap, the type that you would use to fasten the canoe to the roof rack of your care, wrapped around the tree threaded into the buckle, carribener at head of Hammock clipped on and just adjust tension,( see my pic gallery)
I do realise weight is important to some of you Hikers, but I'm sure the weight would not be that much more than a couple of slings.
Happy Hanging Regards Bill
BillyBob58
11-22-2008, 18:02
I'm considering using the mini's on my HH ULBA. Bringing this up again to see what the testers said. (I couldn't find the consolidated results.)
How'd they do?
Mine worked great, at the cost of a couple of oz. Though I have simplified and gone back to original HH hitch lately, I remember it is much faster than the HH fig 8 hitch. Particularly if you needed to make and adjustment height wise or centering wise. And you can apply enough tension such that you need to be careful to avoid stretching or breaking something. Due to concerns about the 250 lb rating, I would always take some of the excess rope and tie a back up knot, just enough to minimize speed of fall in case of failure. This concern was probably unwarranted considering the 250 lbs(?) was a working load, not a breaking load.
I think I remember another problem was having to reposition them each set up, depending on the distance between trees. This decreased their convenience factor somewhat. I need to try them again.
Just be careful and don't get aggressive when tightening, or you might break or stretch your ridgeline. Another question is whether they will fit all HH ropes. My Explorer UL ropes fit perfectly, but my friend questioned if his Expedition ropes would fit, as they were thicker.
Awesome thread. I was expecting the loads to be much higher, but in the end the math is pretty straightforward and makes sense. I have a degree in mechanical engineering, but never put this problem to paper.
A few years ago I made A-frames to support a Brazilian style hammock in the back yard. The lines were extended over the A-frames and then attached to stakes in the ground, but it never held. The stakes always pulled out or bent.
After playing with my new HH Explorer for the first time, I decided to order the Hitchcraft devices (the double pack, 2-mini + 2- monster).
As a sailor I appreciate the concept of running rigging. I'm sure these babies will make hanging the HH much easier.
I'm not sure of the diameter on the HH suspension, but the Monster RopeTie from Hitchcraft is perfect on the Clark NA. Needs no safety knot or hitch to keep it from slipping, and can be used exactly as shown on the Hitchcraft instructions.
The 3mm Amsteel with the mini ropeties does need an extra safety hitch to keep it from slipping - just slightly too small and too slippery (this is the rope used on the BlackBird suspension.) But I love the hitchcraft devices - super simple, and easy on the rope. And fairly light for their size. They are on all my hammocks with rope suspension adjustments.
As with any new device on a hammock, check and double check before putting your entire weight (and trust) in it. Have a crash pad below you the first time in case the device slips. If it does, just make a half hitch around the cleat on the device with the tail end of your rope.
swampfox
04-05-2009, 07:59
I don't know how in the world I have missed this thread, but I have just gone through every page. More answers lead to more questions so now I'm confused at a higher level about more important things.
The one thing I'm sure of, I gotta get me some of these!
hacktorious
04-05-2009, 14:47
4mm seems to work the best. 5mm for some reason slips. I never had an issue with 4mm. Now that I have switched to 5mm, I have some minor issues with it.