Find Length of Pulley Belt

greatwhiteshark

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The drive wheel of an engine is 13 inches in diameter and the pulley on the rotary pump is 5 inches in diameter. If the shafts of the drive wheel and the pulley are two feet apart, what length of belt is required to join them?
 
greatwhiteshark said:
The drive wheel of an engine is 13 inches in diameter and the pulley on the rotary pump is 5 inches in diameter. If the shafts of the drive wheel and the pulley are two feet apart, what length of belt is required to join them?

What have you done so far to solve this, Janet?

Draw 2 representative circles a bit like this: O o (diameters 13 and 5 inches)
Make a line joining their centers: that line = 24 inches.
Draw a representative belt:
the belt's length will be 1/2 the circumference of each circle plus the 2 straight lines formed by the belt at top and bottom.

See what you can do...
 
tkhunny said:

Yes.

What's wrong with this:
1/2 the circumference of each circle plus the 2 straight lines formed by the belt at top and bottom. ?

(circumference + circumference)/2 + 2(straight line)

pi(13 + 5) / 2 + 2sqrt(24^2 + 4^2) : right?
 
I don't think so. It's greater than ½ on the large pulley and less than ½ on the smaller pulley. The two free sides of the belt aren't parallel.
 
tkhunny said:
I don't think so. It's greater than ½ on the large pulley and less than ½ on the smaller pulley. The two fee sides of the belt aren't parallel.
Yikes! You're right of course...my 1st mistake...to-day!

Well, who knows; perhaps doing it my way is the origin of the expression
"tighten your belt"...
 
To partly atone for my sin:
length of each "straight belt portions" = 23.66~

I used an isosceles triangle (big circle at bottom, smaller near top)
with the equal sides being tangents to both circles.

The 23.66~ is the distance between the tangent points, of course.

Distance from top of small circle to top of triangle = 12.5

NO, I don't care to elaborate or finish this off!
 
Well, had some time to kill between reciting Holy Rosaries:

u = 6.5 (radius larger wheel)
v = 2.5 (radius smaller wheel)
w = 15 (distance between whells, from rims)
x = straight portion of belt

x = sqrt[(k + w + u + 2v)^2 - u^2] - y
where:
k = v(2v + w) / (u - v)
y = sqrt[k(k + 2v)]
That works out to x = 23.664319132...

Now, calculate angle A formed at centers of wheels, by the lines from center
to tangent points: that'll be A = 160.811863546... degrees.
Use this angle to calculate belt portion on wheels:

smaller wheel a = pi * v * A / 180 ; that'll be a = 7.016741236...
larger wheel b = pi * u * (180 - A) / 180 ; that'll be b = 22.597177278...

So belt length = 2x + a + b = 76.942556778...

Hope someone checks this!
 
Are you SURE the belts go around the outside? :wink: Maybe they cross each other in the middle. It's not an uncommon design for deliberately increasing belt friction. If only the problem statement had clarified this point.
 
I agree with Denis' answer.
I drew lines perpendicular to belt and a line from the center of small circ parallel to belt. That makes a 4, L, 24 right triangle. The small angle T satisfies
sin(T) = 4/24
T = .167481
L = 24cos(T)
Small belt arc =
2pi-2(T+pi/2) = pi-2T
Big belt arc = pi+2T
Total = 2*24cos(T)+2.5(pi-2T)+6.5(pi+2T) =
48cos(T) + 9pi +8T
Same answer, different method.
 
First, thanks to all of you for not "picking" on me on the
over-complicated looks of my solution up there....:)

Hope to be forgiven with this:

u = 6.5 : larger radius
v = 2.5 : smaller radius
w = 24 : distance between centers
x = one straight portion of belt
A = angle determining arc length
y = half-portion on small wheel
z = half-portion on large wheel

x = sqrt[w^2 - (u - v)^2] : 23.66~
A = acos[(u - v) / w] : 80.41~ degrees
y = (pi * v * A) / 180 : 3.51~
z = [pi * u * (180 - A)] / 180 : 11.30~

belt = 2(x + y + z) : 76.94~

NOTE: belt does not cross between wheels :)
 
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