nemo Posted February 17, 2006 Report Share Posted February 17, 2006 So I'm rolling up a tie-down strap (the ratcheting kind - Irish's got me moving appliances today), and I'm watching my quarter-turns get more and more productive as the roll becomes bigger. Now I'm curious - what is the formula for determining how productive each quarter turn will be? Assuming that the strap is 1mm thick, how do I know how much strap I will wind on my 47th turn, as opposed to my second? Thanks,Nemo Quote Link to comment Share on other sites More sharing options...
Tormod Posted February 17, 2006 Report Share Posted February 17, 2006 Hey Nemo! ;) Wouldn't we also need to know the initial dimensions of the roll when it is completely unwound? Or at least the dimensions of the first complete turn of the strap. After that I guess this is simply a spiral, in which case this might be helpful: http://www.newton.dep.anl.gov/askasci/math99/math99015.htm Quote Link to comment Share on other sites More sharing options...
CraigD Posted February 17, 2006 Report Share Posted February 17, 2006 …how do I know how much strap I will wind on my 47th turn, as opposed to my second?The length of strap pulled in 1 complete turn of the drum is2*Pi*R, where R is the radius of the drum including the thickness of the strap wound around it. The radius of the drum increases by 1 thickness of the strap after each complete turn. Since the radius of the drum is an arithmetic progression, we can write total length of strap pulled in N complete turns asL = 2*Pi*((R0 +(R0 +N*H))/2)*N = (Pi*H)*N^2 +(2*Pi*R0)*N, where R0 is the initial radius of the drum, and H is the thickness of the belt. Looks a lot like the equation for an object under constant accelerationChange in distance = (acceleration/2)*Time^2 +(initial velocity)*Time, eh? Although this equation is fine for whole number values of N – complete turns, you have to fudge it a bit to get fractional turns, since the R doesn’t increase continuously, but in discrete jumps. Just calculate L for a whole number of turns N, then add2*Pi*(R0 +N*H)*F, where F is the additional fractional of a turn. So for a .01 meter drum (seems about the size I remember the little ones being) with .001 meter thick strap, the 2nd pull would pull about 16 mm, while the 47th would pull about 33. Although real world belts stretch and slip, I think this equation would agree reasonably well with observation, so long as tension of the belt is constant. So, it’s really only good for describing tie-downs used for stuff like pulling pallets over concrete, not the usual job of tying down loads ;) IrishEyes 1 Quote Link to comment Share on other sites More sharing options...
IrishEyes Posted February 18, 2006 Report Share Posted February 18, 2006 Do you see what I live with?I don't know of a single other person in the world that would have that question while trying to bring our fridge to our new house. Geez, I love that man!! Quote Link to comment Share on other sites More sharing options...
nemo Posted February 19, 2006 Author Report Share Posted February 19, 2006 Looks a lot like the equation for an object under constant accelerationChange in distance = (acceleration/2)*Time^2 +(initial velocity)*Time, eh?Mmm... yeah - that's exactly what I was thinking:hihi: Awesome answer - thanks. Quote Link to comment Share on other sites More sharing options...
infamous Posted February 19, 2006 Report Share Posted February 19, 2006 Geez, I love that man!!Ahhh, to be loved; A fortunate thing indeed.............Nemo...count your lucky stars................Infy Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 20, 2006 Report Share Posted February 20, 2006 I think Craig you're missing the term +N/2 from the arithmetic progression sum, which is N(N + 1)/2. Quote Link to comment Share on other sites More sharing options...
CraigD Posted February 20, 2006 Report Share Posted February 20, 2006 I think Craig you're missing the term +N/2 from the arithmetic progression sum, which is N(N + 1)/2.No, I included it. It canceled nicely with the 2*Pi*R equation for circumference of a circle, so no “2” appears in the final equation(Pi*H)*N^2 +(2*Pi*R0)*N I did leave out the 2*Pi term in the equation for the length of strap pulled by the final fraction of a turn. I’ve corrected it now. I've suffered from sloppiness my whole life! :lol: Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 20, 2006 Report Share Posted February 20, 2006 Hmmmmmmmm.... 2*pi(R*N + H*(N + 1)*N/2) is what I got. Quote Link to comment Share on other sites More sharing options...
CraigD Posted February 20, 2006 Report Share Posted February 20, 2006 Hmmmmmmmm.... 2*pi(R*N + H*(N + 1)*N/2) is what I got.OK, attempting to get this right, at last…Initial radius: R0Final radius: R0 + H*N, where H is thickness of strap, N number of revolutions of the drumSum of R0 + R0+H*1 + … R0+H*(N-1) = N*((R0) + (R0 +H*(N -1)))/2 = (H*N^2 +(2*R0 –H)*N)/2(not (H*N^2 +(2*R0)*N)/2, as in original post!) So, substituting for R in diameter of circle L= 2*Pi*R,L= (Pi*H)*N^2 +(Pi*(2*R0 –H))*N So I left out the “-H” term. Sloppy, sloppy, sloppy! Quote Link to comment Share on other sites More sharing options...
arkain101 Posted February 21, 2006 Report Share Posted February 21, 2006 Watch the end of the strap when it moves on the floor. Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted February 21, 2006 Report Share Posted February 21, 2006 Gosh Craig you do things complicated!!!!!:)Sum of R0 + R0+H*1 + … R0+H*(N-1) = N*((R0) + (R0 +H*(N -1)))/2 = (H*N^2 +(2*R0 –H)*N)/2I can see now that you're counting the first wind as radius R, not R + H. We're both approximating, of course, and neither of us has been bothering about what happens at the end of each wind!:)Take my last expression and subtract pi*N*H, or add it to yours, I s'pose, and then add epsilon at the end of each wind. :) Quote Link to comment Share on other sites More sharing options...
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