Comments on: The Involuting Goat

By: Hex

Hex — Fri, 06 Aug 2010 14:24:45 +0000

@newguy4321, This issue bothered me at first when I saw Bilbao’s solution. But since the green lines and the blue lines are always perpendicular (radius vs tangent), it is ok.

By: newguy4321

newguy4321 — Wed, 04 Aug 2010 15:56:47 +0000

First off I want to say great site and great question! I was looking over Bilbao’s solution and I had just one question. I am having trouble seeing how the angle dt in the blue circular region is directly equal to the dt in the green circular region. In the green region the angle dt is the angle formed between two tangent lines of the circle. Am I missing some geometrical relationship here? Thanks for your time!

By: james turner

james turner — Sat, 23 Jan 2010 19:51:37 +0000

Several answers are very close, which is surprising given that the integration approaches yield exact solutions. This leads to a simple question: why are there differences? I believe that the differences can be traced to the curve sweep out from the tangent line to the circle. For example, Bilbao assumes that his polar differential area uses a triangle height that is measured tangent to the circle, whereas Turner uses a differential area where the height is parallel to the tangent line. As a result, the coordinates of the desired curve are given by:
Bilbao: r{cos(phi)-(theta-phi)*sin(phi), sin(phi)+(theta-phi)*cos(phi) }
Turner: r{cos(phi), sin(phi)+theta-phi}
where theta is the unknown max. angle and phi is variable.
The start and end points are the same. The y-axis components of the curve are different. The difference is small and accounts for the variation in answers. All this means is that the problem definition is vague, in the sense that it is unclear whether the “side” areas are to be minimized or maximized?

By: bilbao

bilbao — Mon, 18 Jan 2010 08:41:58 +0000

A few little comments:

1. First of all, thanks hex for your detailed explanations

2. Congrats to everybody who tried the puzzle…

3. I remember learning at university how to design gears, and the very basis had to do with drawing involute and evolute curves to shape the gear teeth; the very same curve as in this puzzle, thus the title

4. At least with difficult puzzles I try to give a little clue in the title: ‘involuting goat’, ‘complex clock’,…these titles were not by chance…

5. ‘When there is a will, there is a way’. Wow, what a beautiful way to summarize how I face any puzzle or real-life problems.
Example: let’s suppose you face this problem with no knowledge about math software and you get to a very difficult equation…never give up…first feel satisfied to have reached this point…then, since there is always a way, look for alternatives…either ask yourself if a software may exist…or use excel!!…by using as many columns as you need you can build and solve almost any equation, piece by piece, and leave just one first cell to enter a value for the variable. You can get very exact solutions this way…
I always get deeper satisfaction whenever I manage to overcome an obstacle towards the solution than getting the exact solution itself…

By: hex

hex — Sat, 16 Jan 2010 20:39:14 +0000

Suineg, I too had not realized that the term involute describes the basic problem of this puzzle. It was only after I had a look at Bobo The Bear’s link and researched further that I found out it was a common issue in mechanical engineering, especially in involute gears where the gear tooth shape is derived similarly.

Check out http://en.wikipedia.org/wiki/Involute_gear and http://science.howstuffworks.com/gear8.htm for more info.

Another application is involute turbines.

As for intersection area of 2 circles, check out the encircled goat puzzle http://www.smart-kit.com/s2231.....at-puzzle/

By: suineg

suineg — Sat, 16 Jan 2010 00:36:17 +0000

Thanks Hex, I missed that important detail, when reality meet maths, cool, the answer were pretty stunning man, However the intersection of the circles exercise geometrically was really hard to solve even without that detail, for me at the end was unsolvale algebraically.
Bilbao: You know what was really cool about this puzzle, that you give a clue in the title man and I did not even realized that, after Hex clarification I realized that this problem was an all around hard problem both mathematically and generally speaking so I take back my comment about difficulty been too focused in the mathematical aspect, but cool this help me read 2 books of trigonometry to learn some of what I missed back then.

By: hex

hex — Fri, 15 Jan 2010 22:19:39 +0000

The difference between the 3 solutions pertains to the involute part and is as follows:

Bilbao has divided the area into triangles and has summed them. By making these triangles infinitesimal, the sum turned into an integral. Equating the areas yields a 3rd degree polynomial/equation which could have been solved algebraically (too difficult but feasible), or by using math software as Bilbao did. I like this solution because it uses the fundamentals of calculus, uses a single integration to obtain the involute area, and yet reaches a 3rd degree equation.

Hex has derived the equations of the boundaries of the area and integrated using cartesian coordinates. Equating the areas yields a crazy equation with sines/cosines, practically unsolvable except by using math software. Using polar coordinates instead would have yielded eventually the same equation as Bilbao’s.

BoboTheBear used the results of the same puzzle in Wolfram, with a few modifications.

By: hex

hex — Fri, 15 Jan 2010 21:47:51 +0000

Suineg, you have to take into consideration that the rope curves along with the barn. In other words, the goat curve is not a circular arc.

By: suineg

suineg — Fri, 15 Jan 2010 16:00:27 +0000

Cool answers man, my approach was different from the one taken by Bilbao, Hex and Bobo the Bear that seems pretty similar, Polar cordinates gives back time, college, no responsabilities and only oportunities, good memories, anyway, I think that my approach was similar to the one took by Shawn for his comment about many arc cosine and square roots, So for the sake of my mathematical background I have a question for Bilbao is my approach inexact because I dont use integrals if so why, thanks in advance man if you can clarify me that doubt, cool.

By: RK

RK — Fri, 15 Jan 2010 14:18:58 +0000

Here is Bilbao’s solution:
http://www.smart-kit.com/wp-co.....ution1.jpg

By: hex

hex — Fri, 15 Jan 2010 11:41:25 +0000

Bobo The Bear’s link shows 2 things regarding my approach to solving this puzzle:

1- I wonder how I missed the method of calculating the area using polar coordinates (1/2 integral(r^2 dt)) which leads to a very simple solution without the need for math analysis software

2- When there is a will, there is a way!

#2 is way more important than #1

Bilbao, I researched the word involute and found out that it has several uses in your field. And I was wondering from where you coined the term hahaha

By: Bobo The Bear

Bobo The Bear — Wed, 13 Jan 2010 20:54:21 +0000

I found a description of this problem (as well as the previous gaot problem) at the following site:

http://mathworld.wolfram.com/GoatProblem.html

Using the information found there, I came up with a value of approximately 15.07882679m for the lenght of the rope. I hope this agrees with your results. Solving a cubic equation for L (the length of the rope) would yield an exact answer, but I have not done this yet.

Thanks for keeping our math skills sharp, bilbao!

By: suineg

suineg — Wed, 13 Jan 2010 20:25:02 +0000

CBD and CAD should be multiplied by pi/180, missed that step at the end sorry

By: suineg

suineg — Wed, 13 Jan 2010 18:38:34 +0000

Ok Bilbao and hard equals trigonometry.

The problem itself is simple: You have the area of the barn: 144pi
then you have two circles:
Circle 1 that is the barn and does not change
Circle 2 that is the one formed with the rope attach to the goat, this one is variable for this problem

So now what you have to do is find a radius for Circle 2 so that the Area of Circle 1 is equal to the Area of Circle2 minus the Area of the intersection of this two Circles (sound easy)
the equation for this: AreaC1 = AreaC2 -AreaC1InterC2 this is:
(1)144 * pi = R2* pi – AreaC1InterC2- (this one is so hard to find)
In fact what you have to do next is put this intersection Area in function to R2 so that you can solved for this value:
For this you have to use a lot of trigonometry (man why man why!!)
For this Circle 1 has r0,the center is labeled A
Circle 2 has r1,the center is labeled B
C and D are the intersection points and c is the distance between A and B in our case is always r0

First you use the cosine formula ( my gosh):
r0^2 = r1^2 + c^2 – 2*r1*c*cos(CBA)
cos(CBA) = (r1^2 + c^2 – r0^2)/(2*r1*c)

Having found CBA, then (2) CBD = 2(CBA).

Similarly,

cos(CAB) = (r0^2 + c^2 – r1^2)/(2*r0*c)

and then (3) CAD = 2(CAB)

Express CBD and CAD in radian measure by multiply bi pi/180. Then we find the segment
of each of the circles cut off by the chord CD, by taking the area of
the sector of the circle BCD and subtracting the area of triangle BCD.
Similarly we find the area of the sector ACD and subtract the area of
triangle ACD.

AreaC1InterC2 = (1/2)(CBD)r1^2 – (1/2)r1^2*sin(CBD)
+ (1/2)(CAD)r0^2 – (1/2)r0^2*sin(CAD)

that was the area of intersection (WOW as hard as it can get)
Now put that in equation (1):

144pi = r1^2*pi – (1/2)(CBD)r1^2 – (1/2)r1^2*sin(CBD)
+ (1/2)(CAD)r0^2 – (1/2)r0^2*sin(CAD)

Now from (3) and (4) and knowing that c equals r0=12 you get:

144pi = r1^2*pi – (1/2)(2*acos(r1/24))r1^2 – (1/2)r1^2*sin(2*acos(r1/24)))
+ (1/2)(2*acos(288-r1^2)/288)*144 – (1/2)144*sin(2*acos(288-r1^2)/288)

This could be simplified more: sin(acosine(x)=(1- x^2)^(-1/2)
but I left it like that because acosine(x) is unsolvable manually

All is in function to r1 now you found the value, this is impossible algebraically at least for me and I read the comment of using maple so this is as far as I will go.
For me the difficulty in this problem relies to heavily in the knowlegde of trigonometry and the use of mathematical software and poorly on problem solving abilities, but cool man cool. Anyway there is an algebraic way of getting the value of arcosine(x) I mean manually of course, cool

By: Shawn

Shawn — Wed, 13 Jan 2010 14:53:00 +0000

Oops, forgot to carry the one!

15.1069m