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# Challenging geometry problem, involving discs (gears, pulleys).?

If I have two discs, diameters 6 and 18, that are touching (tangent), what would be the shortest rope I could put around them both?

Is it possible to get explanations? Without them, I cannot determine who is correct. i used to be able to do this, but my mind is blank right now.

### 10 Answers

- Anonymous1 decade agoFavorite Answer
Suppose the center of the small pulley is at (0,0) and the center of the large one is (12,0). The radius of the small is 3 and that of the large is 9, so if they're touching, their centers are 12 units apart.

The rope will go less than halfway around the small pulley, more than halfway around the large pulley, and be tangent to both of them. We need to find the points that are tangent to both circles and the distance between them.

Draw a line tangent to both circles, and from the tangent points to each of the centers. Because a tangent is perpendicular to each radius, these two line segments (through the radii) are parallel. You also already know their lengths, 3 and 9. Now draw a line parallel to the x-axis that goes through the tangent point of the small circle.

What you've got is a right triangle. The hypotenuse has length 12 (because the opposite sides of the parallelogram are equal and the length between the centers is 12). The leg of the right triangle that goes toward the center of the large circle has a length of 6 units (nine for the large circle minus three from the small one).

A right triangle with a leg half the hypotenuse? This problem gets easier once you realize this. It's a 30°-60°-90°!

The tangent segment, the other leg of the triangle, is 6√3 units long. Furthermore, the points on the circles where the tangent lines meet are 30° off perpendicular. This means it travels 2(90° - 30°), or 120° around the small pulley, and 2(90° + 30°), or 240° around the large one.

The rope goes 1/3 of the way around the small pulley, 2/3 of the way around the large pulley, and in each of the two tangents between them, 6√3 units.

C[small] = πD = 6π. 1/3 of this is 2π.

C[large] = πD = 18π. 2/3 of this is 12π.

The rope must be 2π + 12π + 2(6√3), or

(14π + 12√3) units in length, or

approx. 64.7669 units.

- 1 decade ago
If the diameters are 6 and 18, then the radii are 3 and 9, and the distance between their centers is 12.

If you draw a tangent segment to both circles (the endpoints being the points of tangency), then you've formed a quadrilateral, with the tangent segment, a radius in each circle drawn to the point of tangency, and the segment joining the centers.

You can divide this quadrilateral into a rectangle and a right triangle, and use the pythagorean theorem to determine the length of the tangent segment, x:

x^2 + (9 - 3)^2 = 12^2

x^2 + 6^2 = 12^2

x^2 = 144 - 36 = 108

x = 6sqrt(3)

So part of the rope will be two of these lengths. The next step is to determine what part of the circumferences of each circle will make up the rest of the rope. This is difficult to show in pure text, but it involves trig. If A is the angle formed by the segment joining the centers and the radius of the larger circle drawn to the point of tangency from before, then

cos A = (9 - 3)/12 = 1/2, and

A = 60 degrees.

So the arc of the larger circle covered by the rope is

360 - 2(60) = 240 degrees

which makes that arclength

(240/360)(2π*9) = (2/3)(18π) = 12π

That also means that the arc of the smaller circle that is covered by the rope is

360 - 240 = 120 degrees,

and THAT arclength is

(120/360)(2π*3) = (1/3)(6π) = 2π

SO... we have a grand total of

2(6*sqrt(3)) + 12π + 2π ≈ 64.767

- 1 decade ago
I had some help with this, I'm 12 and a math wiz.

Circumference of Circle A=9.42 units

Circumference of Circle B=28.26 units.

The distance across both circles (from center to center) is 12 units, by adding the radii of both circles, forming the base of an imaginary triangle. The other leg of that triangle has a measure of six units. Assume the line XY of triangle XYZ forms the hypontenuse of Tri. XYZ. The measure of Line XY is about 13.41 units. The final figure would be, rounded to the nearest hunderth:

[(6)3.14/2]+[(18)3.14/2]+(13.41*2)=9.42+28.26+26.82=...

64.50 Units of rope!

Source(s): Some trig and calculus book. - SteveLv 71 decade ago
This is the classic belt length problem and the solution is as follows:

Let A be the angle the rope makes with the line between centers.

Then tan A = (R-r)/(R+r) ; Solve for A in radians.

The rope length not touching the discs is 2 * (R+r) *cos A

The rope length around the smaller disc is (pi - 2A)*r

The rope length around the larger disc is (pi + 2A)*R

Add'em up for the total minimum rope length

Source(s): A head for geometry - How do you think about the answers? You can sign in to vote the answer.
- Anonymous1 decade ago
If they are not in the same plane but instead are positioned side by side then the shortest rope is 18pi.

- 1 decade ago
12(pi) + 12 x (the square root of 5), which is about 64.53192757 . Don't forget your units.

- Anonymous1 decade ago
64.7669, or to significant digits, 65. I solved this graphically rather than with calculus, though if you need the calculus procedure, I can probably dig it up for you.

- 1 decade ago
my answer would be 6sqrt(3)+16pi... As for how i get this answer, it involves some geometry and you need to draw it out and use some trick on this...