# Thread: How much Gross Grain is left on the roll

1. Originally Posted by rjcress
Uh...
was any of that in English?

It must be, I recognized the word DIY.
I can't even spell DIY, let alone tell what they were saying.

2. Spoken like a true professor, Grizz...

Took me a sec, but all my college math came back... Now if I had to figure that out, I would have had to write a differential equation... then my head would have exploded...

3. I use one of these..? Seems to work.

hyotaeemoforggl's

(hike your own trigonomic algebraic estimating equation methods of figuring out remaining grosgrain lengths)

4. math can solve many problems but there seems to be a missing of the simple solutions at times and the formulas dont take into account some of the variables like tightness of the winding and other things like variable thickness of material.

5. Originally Posted by wildcrafter
...the formulas dont take into account some of the variables like tightness of the winding and other things like variable thickness of material.
But WV's solution does... weigh the full roll of known length to find out what one foot or one yard weighs, then weigh it as you go to find out what length remains.

6. All these linear methods are fine and dandy for the WW rolls. But try incorporating them when you are using a bulk spool of multiple wraps, or a cone shaped spool used for industrial purposes.

It's easy really...

Write down the stated amount of product on the reel or cone. Then subtract each unit from the total as you use it. That's how the pros do it.

Or... develop a good eye. You don't need to know how much is on the reel. You simply need to answer one question. To wit..."Is there enough?" If yes... then use it. If not or unknown.. buy some more. No need to get all mathlete.

7. Those sure are some funny looking inch marks on your tape measure, G.

Originally Posted by gargoyle
I use one of these..? Seems to work.

hyotaeemoforggl's

(hike your own trigonomic algebraic estimating equation methods of figuring out remaining grosgrain lengths)

8. Originally Posted by MacEntyre
Originally Posted by wildcrafter
math can solve many problems but there seems to be a missing of the simple solutions at times and the formulas dont take into account some of the variables like tightness of the winding and other things like variable thickness of material.
But WV's solution does... weigh the full roll of known length to find out what one foot or one yard weighs, then weigh it as you go to find out what length remains.
You can also work backwards: weigh a short length and calculate from that. It's surprisingly accurate.

Originally Posted by Ramblinrev
All these linear methods are fine and dandy for the WW rolls. But try incorporating them when you are using a bulk spool of multiple wraps, or a cone shaped spool used for industrial purposes. True. If there's a cone or a spool, you need to know what itweighs, so when you empty one, weigh it, and suntract from the weight of a new, full spool...

It's easy really...

Write down the stated amount of product on the reel or cone. Then subtract each unit from the total as you use it. That's how the pros do it.
Fine, as long as you don't forget ...

Or... develop a good eye. You don't need to know how much is on the reel. You simply need to answer one question. To wit..."Is there enough?" If yes... then use it. If not or unknown.. buy some more. No need to get all mathlete. This is a fine solution, and one I frequently use, but it deprives me of the fun of doing the math.

9. As long as you don't forget what?

The second says exactly the same thing, but it seems odd to me to create D^2 - d^2 by the product (D-d)*(D+d) and then get rid of the extra 2 from D^2 = (2r)^2 with two separate divisions by 2. Huh.
I find that second one intuitive if you think of 'N' as the approximate number of turns left on the roll. Solve for N first, etc...

I love the smell of numbers in the morning...