1. ## Cat Cut Cuestion

In the cat cut excel spreadsheet that BlackBishop offers on his BlackCat tarp site (thanks to Ivelph) when I put in my measurements I get all my numbers as expected, but what I don't understand is that the spread she starts at 3/8" and then ends at 1/4" Wouldn't this make the cat cut kind of "off kilter"
Please explain why it is like this and how you make a perfect cat cut with these measurements.

Thanks

2. The problem is that when converting to 1/8" intervals there is some round off error. The measurements giving are a guide. You won't be able to cut or sew it perfectly anyway, so it is nothing to worry about. Besides, it is only 1/8" we are talking about, and I don't think we will crash into Mars. If you want the cut to be perfect use the first half measurements and then flip your measuring tape for the second half. I still doubt it will be perfect, because I rounded in other areas too. The actual measurements are transcendental and therefore are impossible to get exact (not to mention the statistical impossibility of having exact measurements).

Edit: Also, the last measurement is actually 5/8" from the edge while the first measurement is 1" from the edge. This is another reason for the difference near the edges.

3. Originally Posted by lvleph
You won't be able to cut or sew it perfectly anyway, so it is nothing to worry about.
Exactly what I was thinking when I read the orginal post.

I had the same concerns on my first attempt, but just kept reading the part where Blackbishop said to just smooth out the line by eye and you'll be fine. I'm more than certain that my BlackCat is more than 1/8" off end to end, but unless you put a tape measure to it, you'd never know it looking at it. You'd be too easy asking me things like "who in the world taught you how to sew?"

Good luck with it cameronjreed, of all the things I've made this is the thing I'm most proud of and it is far from perfect.

4. Mine is perfect, but that is because I am a bad ***. JK

However, I cannot believe how taut that thing can pitch. And it stays taut through some serious rain.

5. Well, I'm certainly no bad ***, but I think I must be lucky ! The only tarp I've had any real problems getting taut is the stock HH fly. However, the MacCat/BlackCat design suuuurrrrre does make it easy!

6. ## cat curve formula

The question cameronjreed asked peaked my interest and so I got the spreadsheet and had a look.

First of all, you guys that use this have no idea of the sophistication that is lurking under the covers. IvLeph clearly put a lot of effort into both the calculations and the presentations part. Nice job!

But round-off error is something I know about, and there is more than round-off error going on here. I expected (as did cameronjreed) that the function would be "symmetric" (ain't that right CJR?), by which is meant that if you flip the graph around the center-line axis (x=72/2 in the example case), then nothing changes. CJR was looking at the initial points. If the function was symmetric, then the value 4 inches from the left (1.25") would be the same as the value 4 inches from the right (1.75"). Now a difference of 0.5" relative to 1.25 or 1.75 is really pretty big, even when you're rounding to 0.125".

EDIT : the spreadsheet doesn't make the error that I thought it did, described below. The non-symmetry of the curve is still much larger than I would expect due to round-off, but I've not an answer for that now. Maybe I'll try this same calculation in a different language and see what emerges there.

So I dug a little deeper to see just exactly what function is being plotted here (parabolas I know and love, catenary curves are new to me). Well, the formula is

y = H - alpha*( exp(x) + exp(-x) - 1 )

where exp(x) is the natural number e raised to power x. The thing is, this function is not symmetric over the interval [0,72], e.g., if c = the center point of the x domain (e.g., 72/2 = 36) then if you compute the function value s units to the right of c it is

H - alpha*( exp(c+s) + exp( -c -s ) - 1 )

and if you compute the function value s units to the left of c it is

H - alpha*( exp(c-s) + exp( -c + s ) - 1 )

and these values aren't the same. Round-off or not.

But the catenary curve is symmetric about its axis, and so something is amiss here. What is amiss is that c needs to be 0.

So I'm thinking Ivleph that your loop that computes the values out needs to run from x= -L/2 to x=L/2 rather than x=1 to L.
Then the center point of the curve is relabeled to be 0 rather than L/2, and all is well in the universe. Rather, you can run x over the same interval (because you use that in the output column) but instead you should compute

H - alpha*cosh( x - L/2 )

Not sure what you're solving for with the Newton method, so can't really tell if this offset affects it to. Guessing not, as it seems the iterative solution is working out the necessary relationships between some of the parameters in the model.

Just as I experienced with cutting a curve for the bridge hammock, it seems that getting the details "exactly right" don't matter too much for functionality of the tarp, and the variance introduced by sewing is of the same scale as the deviations from symmetry. But thought you'd want to know.... (along with the rest of the Internet universe )

Grizz

7. Alpha is the parameter being solved for in Newton's Method, it is what makes the catenary curve. Alpha has to be solved using Newton's Method because the catenary function is nonlinear.

The center is not 72/4, in the example given, it is actually (74 5/8)/2. The given measurements are every inch. That is why it appears not to be symmetric.

If you look closely at my loop you will notice a change of variable resulting in the loop ranging from -a/2 to a/2. At least that is what my intention was. It appears that is what I have done from what I can tell by quickly looking over the code just now. If that is not the case let me know what I have done wrong and I will try to fix it.

8. Originally Posted by lvleph
Alpha is the parameter being solved for in Newton's Method, it is what makes the catenary curve. Alpha has to be solved using Newton's Method because the catenary function is nonlinear.

The center is not 72/4, in the example given, it is actually (74 5/8)/2. The given measurements are every inch. That is why it appears not to be symmetric.

If you look closely at my loop you will notice a change of variable resulting in the loop ranging from -a/2 to a/2. At least that is what my intention was. It appears that is what I have done from what I can tell by quickly looking over the code just now. If that is not the case let me know what I have done wrong and I will try to fix it.
Thanks, I understand now why it "looks" asymmetric. Sorry for the fire alarm!

Grizz

9. That section of the code was a pain. It took me a while to even understand how to get it to work.

10. Just wanted to thank everyone who chimed in. Especially Ivleph and Griz. I am not a math wiz like you guys, but I am of the opinion that every little bit of exposure to something that you are not good at helps improve that skill.

I understand that cutting and sewing will cause imperfections that are outside of anyones control, but I was just curious as to why the cat cut was not symetric about the axis. Now I understand.

Thanks again. Off to make my first cat cut pattern.
BTW... If anyone is interested....If you live near the headquarters of your local newspaper they sometimes have long, wide rolls of blank paper called "end rolls" that are GREAT for making patterns. Fortunately for me I also have a buddy who works at the dynomite factory. They have the same type of thing (large wide roll of paper) but it is much thicker and IMO easier to work with.
Just and FYI.

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