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Distance between 2 pullies 
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#1
Mar2910, 07:26 AM

P: 21

hi there,
i have two pulleys of known radii. i also have a belt of known length (radii). I need to find the distance between the two pulley centres so that the belt raps neatly around both pulleys. I've had a go at it for a while now and the furthest i have come is: 1) to make an equation that is not solvable (i have an angle inside and outside a cosinus function) & 2) to start the problem off with the distance between the pulleys being the sum of the radii then working out what the belt length should be for that distance, then subttracting this length from the actual belt length which gives me the length that is left over on the belt. However when i work backwards from here, substituting a new belt length between the pulleys using the leftover lengths i realise that the angles the tangents(belt) form on the pulleys also change and therefore my whole equation comes crashing down on itself. so, how can i go about calculating this distance? thanks, wernher edit: although this is not a homework question, i believe it falls under the criteria of a textbook type question. sorry for posting this here. 


#2
Mar3010, 01:06 AM

P: 14

Well, I do not know specifics about pulleys so some of my math or assumptions might be wrong, but here goes nothing.
Assume we have pulleys with radii R,r respectively such that R>r. Assume we have a belt with radius p, thus length[itex] P= 2p\pi[/itex]. Let c be the distance from the tops(bottoms) of the pulleys. Then [tex]P=R\pi + r\pi + 2c[/tex] Let x be the distance between the pulleys. Using the Pythagorean theorem, we see that [tex](Rr)^{2}+(R+x+r)^{2}=c^{2}[/tex] so [tex]c=\sqrt{2(R^{2}+r^{2})+2(R+r)x+x^{2}}[/tex] Thus [tex]P=R\pi + r\pi + 2\left (\sqrt{2(R^{2}+r^{2})+2(R+r)x+x^{2}}\right )[/tex] Solving for x, we see that [tex]x=(R+r)+\sqrt{(P  r  r\pi + R  R\pi) (P + r  r\pi  (1 + \pi) R)}[/tex] which is in terms of all our given quantities. Here is a picture I drew that I used. Once again, I do not know about pulleys, so this may all be hogwash. 


#3
Mar3010, 02:05 AM

P: 21

looks good to me. except i don't quite get what [tex]R \pi \ \& \ r \pi[/tex] in the equation of P are? they're supposed to be the length's of the belt on the circumference of the pulley's right?... shouldn't these terms be fractions of the form:
[tex]2 \pi R\frac{180 + 2 \alpha}{360} \ \& \ 2 \pi r\frac{180  2 \alpha}{360}[/tex] ? ( [tex]\left(\alpha\right)[/tex] being the angle of the the hypotomus c, which i worked out was the same angle that lies between the radius extending to the tangents and the vertical radius) 


#4
Mar3010, 10:13 AM

P: 14

Distance between 2 pullies
I assumed that the belt goes from north pole to north pole (south pole to south pole) and thus the length of belt that gets accounted for being in contact with the pulleys would be half of the sum of the circumferences of the pulleys. I do not know enough about pulleys to know if and how to account for the angles



#5
Mar3010, 10:19 AM

P: 21

okay, i guess thats a good assumption(approximation) if the sizes are not that different, what happens though when they are very different and close to each other?



#6
Apr310, 11:56 PM

HW Helper
P: 925

The above solution is approximate as already stated. To get an exact solution, consider the attached picture:
1. The small solid circle on the left has radius r, center C1 = (0,0) and equation y^{2} = r^{2}  x^{2}. The large solid circle on the right has radius R, center C2 = (d,0) and equation y^{2} = R^{2}  (x  d)^{2}. The dashed circle has radius Rr, center C3 = (d,0) and equation y^{2} = (R  r)^{2}  (x  d)^{2}. 2. It should be clear from the geometry that the lower red line is parallel to the upper one, starts at C1 and is tangent to the dashed circle. 3. The lower line has as it's equation y = mx (i.e. y = 0 @ x = 0), where m is the slope of the line. But, m is also the derivative of the dashed circle at the tangent point (x_{t}, y_{t}) 4. By equating m = y/x with the derivative of the dashed circle, you can solve for x_{t} and y_{t}. Using these values, m can be calculated. 5. Once you know m you can find the tangent point on the large circle by equating m with the derivative the the larger circle. The same can be done with the smaller circle. 6. Once the tangent points are determined, you can calculate the "belt length" for any value of d. 7. It might be possible to come up with a single algebraic equation for "belt length" as a function of d, but it would be pretty complicated. If we assume such an algebraic expression L = f(d) can be determined, then to find the value of d, just solve L1 = f(d) for d, where L1 is the desired belt length. 


#7
Apr410, 01:44 PM

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It is my turn!
See the attachment. The hardest part is to see that the angle, a, formed by the separation of the pullies and the difference in the radius is the same as the angle to the point of tangency of the belt on each pully. Once you see that the angle is clearly given by: [tex] Sin (a) = \frac {R r} d [/tex] The lenght of belt between the points of tangency is: [tex] L = d \sqrt { 1  \sin^2 a } [/tex] For the length around the large pully we have: [tex] L = \pi R + 2Ra [/tex] For the small pully we have: [tex] L = \pi r  2ra [/tex] Adding up the pieces gives: Belt Length = [tex] 2 d \sqrt{1 sin^2 a } + \pi R + 2Ra +\pi r  2ra [/tex] 


#8
Apr510, 05:51 AM

P: 21




#9
Apr510, 05:54 AM

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#10
Apr510, 09:22 AM

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[tex] a = sin^{1}( \frac {Rr} d) [/tex] 


#11
Apr510, 12:44 PM

P: 21




#12
Apr510, 01:38 PM

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And we have 2 equations.
a can be completely eliminated from the expression for belt length. So you can get an expression for d in terms of belt length. Unfortunately since d is inside the argument for sine it will not be possible to isolate it. However the expression you get looks to be amenable to a fixed point iteration. That is create an expression for d which has d in it. Then make a guess at d, use it to compute d, take the result and recompute. It should only take a few iterations to find a meaningful value for d. 


#13
Apr510, 01:55 PM

P: 21

iterating would be feasible, but surely there must be some nice equation that describes it ;) 


#14
Apr510, 02:12 PM

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#15
Apr510, 02:17 PM

P: 21

my appreciation of math will come to an end if there is no clean solution. I'm wondering now if it cannot be solved with limits and surface areas. 


#16
Apr510, 02:48 PM

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Well, if this causes you to loose respect for math, then I guess you did not have much respect for it to begin with. You will find that MOST real world applications do not have good closed form solutions.



#17
Apr510, 02:51 PM

P: 21

I'm however a very stubborn person and i will try and solve this until there are no more trees left. edit: by the way when you say most "real world applications" i take it your describing the small details that are left out of equations and which could be included, therefore it would not be the math that fails to describe it but the detail. 


#18
Apr510, 05:35 PM

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We can usually mathematically describe a system in great detail, however this usually results in an equation which can only be solved numerically.
As for your problem I recast the result as: [tex] d = \frac { ( bl  \pi (R+r) 2(Rr) \sin^{1}(\frac {Rr} {d} )} {2 \sqrt{1 ({\frac{Rr} d})^2 } }[/tex] Where bl is the belt length, I used , [tex] d= \frac {Bl  \pi (R+r)} 2 [/tex] the small angle approximation to get a initial guess. By the 2nd iteration the result is more accurate then you will have the capability of measuring. Belt Length 50 Rb 6 Rs 2 D= 12.43362939 This is my first guess 11.74803617 This result would get you a useful separtion 11.74570521 11.74570518 11.74570518 11.74570518 11.74570518 


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