The cow and the silo. - Page 2

Maybeso · 2003-08-21, 00:14

Wackerbarth,

Your description was amazingly easy to follow. Congratulations, I'm very impressed. I expected to have trouble following it even with a diagram in front of me.

To clear up the very last part, you defined Tf as the angle of the chain when the cow hits the x axis. Is this where your integral stops -- to avoid counting the overlap twice?

Consider if the cow continues around. Although the cow and an increasing part of the chain is below the axis, a decreasing amount is above, and sweeps an area of grass that you don't seem to have counted.

To clarify:
At 3 o'clock the chain is horizontally tangent to the 'top' of the silo, and the cow is at (100 - 50*pi/4, 25).
At Tf, the chain has wrapped further around, and the cow is at (x_tf, 0).

Now we install a fence along the +x axis (the cow can still go around).
We also make the chain retractable.

The cow now walks along the fence from x_tf to the silo at 50, the chain retracting to remain straight between the cow and where it is tangent to the silo. Have you counted this area swept by the chain?

I thought at first it was the area of your triangle, but you use the triangle to derive Tf.

P.S. Your description was so good that I could visualize this possible missing piece without a diagram. ;)

[edit] the side of the silo is at 50, not 25. [/edit]

Wacky · 2003-08-21, 02:12

Thanks for catching my error. You are correct, I forgot that part. All of the above answers are wrong.
I left out the area that you indicated.

When the cow reaches the x-axis, there is indeed a small portion that the chain has not yet covered.

I see two ways to approach it. One way is to shorten the chain while at the same time continuing to wrap farther around the silo. The shortening would be such that the cow travels left on the x-axis until it reaches the silo.

The other approach would be to add the area of the triangle that we have omitted and then subtract the piece of the silo that infringes in that triangle.

Taking that first approach, the length of the chain along the tangent to reach the x-axis is given by S*tan(pi - T) or, in our case -25 * tan (T).

To integrate that would require integral (tan^2(T) * dT).

So let's look at the second approach.

When the cow stops at the x-axis, we can form a right triangle. Our chain is the hypotenuse, the x-axis is the base and the height is S * sin (Tf).

Counting both this triangle and its image on the other side of the axis, the triangles add 969.7102 sq ft.

But this is too much. We must subtract the "lens" which is the part of the triangles inside the silo.

2 * ( pi - Tf ) * 1/2 * 25^2 is the area of the part of the silo between the Tf points and the center of the silo. This is 686.4673 sq ft

And finally the triangular part of that wedge gets added back in.
2 * -cos(Tf) * sin(Tf) * 1/2 * 25^2

So the correction is 536.5045 sq ft.

And the revised total is 28016.9129 sq ft

c00ler · 2003-08-21, 09:56

What about 25516.9262122935 sq ft (Using Wackerbarth's Tf)
Or 25516.9260688448 sq ft (Using more precise value of Tf)

The formula is (pi*L^2)/2 + ((L^3)-(L-Tf*R)^3)/(3*R) - (pi*R^2)
Where L - length of the chain
R - radius of the silo
Tf - the solution of Tf - tan(Tf)=L/R

And for short chain(L<=pi*R/2) formula is
(pi*L^2)/2 + ((L^3)-(L-Tf_2*R)^3)/(3*R) - (2*Tf_2*R^2)
Where L - length of the chain
R - radius of the silo
Tf_2 = L/R

c00ler · 2003-08-21, 19:44

Oops. The R*L=2500 sq ft must be added to my answers.

And here is the solution.

L-length of the chain
R-radius of the silo
Coordinate system:
The origin - the center of the silo.
The point of attachment has coordinates (-R,0).

1) if L>pi*R/2

For x<=-R area is (pi*L^2)/2.
Let's see what happens when x>-R (and y>=0)
Imagine another chain with ends at the cow and at the center of the silo (It's made of rubber and it's always straight)
Let's use Wackerbarth's angles T and Tf(To find Tf we solve "Tf - tan (Tf) = L/R" which is very often bruteforce-only (for many values of L/R) ). And another angle A - angle between the rubber chain and the x axis.
While cow is going T is changing from 0 to Tf. And the rubber chain is sweeping some area S. When T=0 the rubber chain, the normal chain and the x axis form a triangle with area of L*R/2=S1.
If we add S1 to S and subtract 1/2 area of the silo( (pi*R^2)/2 ), we'll get the half of area that cow can graze in x>-R area.
So all we need is S.
And now rubber chain helps us:
1)One end of the rubber chain is fixed(in the center of the silo)
2)The rubber chain is straight
3)Other it's end is moving (with a help of the cow)
So
S= integral (1/2)*(length of rubber chain)^2 dA from A0 to A1
Let's look at A-T angle.
We know that length of the chain not wrapped around the silo is (L-T*R)
So tan(A-T)=(L-T*R)/R
Or
A=T+arctan((L-T*R)/R)
And we can calculate length of rubber chain which is equal to sqrt(R^2+(L-T*R)^2)
Back to the integral.
S= integral (1/2)*(R^2+(L-T*R)^2) d(T+arctan((L-T*R)/R)) from A0 to A1
S= integral (1/2)*(R^2+(L-T*R)^2)*(1-1/(1+((L-T*R)/R)^2) dT from 0 to Tf
S= integral (1/2)*(R^2+(L-T*R)^2)*(((L-T*R)/R)^2)/((R^2+(L-T*R)^2)/R^2) dT from 0 to Tf
S= integral (1/2)*((L-T*R)^2) dT from 0 to Tf
S=(1/(2*R))* integral ((L-T*R)^2) d(T*R) from 0 to Tf*R
S=(1/(6*R))*(L^3-(L-Tf*R)^3)
S=(L^3-(L-Tf*R)^3)/(6*R)

And total area is (pi*L^2)/2 + 2*(((L^3)-(L-Tf*R)^3)/(3*R) + L*R/2 - (pi*R^2)/2 )
or
(pi*L^2)/2 + ((L^3)-(L-Tf*R)^3)/(3*R) + L*R - (pi*R^2)
or in that puzzle
28016.9260688448 sq ft

2) if L<=pi*R/2

We just need to do some changes:
1)The cow stops not on x axis but on the edge of the silo.
So Tf is different let's call it Tf_2
Tf_2=L/R
2)The rubber chain doesn't sweep all area of the silo.
So we must subtract only area of Tf_2*R^2

And the answer is
(pi*L^2)/2 + ((L^3)-(L-Tf_2*R)^3)/(3*R) + L*R - (2*Tf_2*R^2)
Where Tf_2=L/R

Quote:

Originally Posted by Wackerbarth

We can integrate 1/2 *(100 - 25 * T)^2 dT through the appropriate angle and have the area covered.

Wackerbarth, can you prove it? Or repeat the proof if I've missed it.

Wacky · 2003-08-21, 21:26

Quote:

Originally Posted by c00ler

Quote:

Originally Posted by Wackerbarth

We can integrate 1/2 *(100 - 25 * T)^2 dT through the appropriate angle and have the area covered.

Wackerbarth, can you prove it? Or repeat the proof if I've missed it.

This is just the formula for area as measured in polar coordinates.
Into that, we have introduced the amount of chain that is sweeping through the grass. ( The 100 ft total minus the amount against the silo. )

I'm not sure what it is that you are asking me to prove?

:arrow: The length of a circular arc is R * Theta
:arrow: The area of a circular wedge is 1/2 * R^2 * Theta

c00ler · 2003-08-22, 08:23

Quote:

Originally Posted by Wackerbarth

This is just the formula for area as measured in polar coordinates.

Yes, but polar coordinate system must be static. But the origin of your system is moving along the edge of the silo

Quote:

Originally Posted by Wackerbarth

The central angle of the silo between the anchor point and point of tangency is T

because the point of tangency is moving.
Or in other words this formula must be used within one polar coordinates system, not within infinite number of them.

Wacky · 2003-08-22, 09:37

Quote:

Originally Posted by c00ler

Quote:

Originally Posted by Wackerbarth

This is just the formula for area as measured in polar coordinates.

Yes, but polar coordinate system must be static. But the origin of your system is moving along the edge of the silo

Quote:

Originally Posted by Wackerbarth

The central angle of the silo between the anchor point and point of tangency is T

because the point of tangency is moving.
Or in other words this formula must be used within one polar coordinates system, not within infinite number of them.

It doesn't have to be static when you integrate. Each incremental piece of area is in its own "static" coordinate system.

c00ler · 2003-08-22, 10:38

Quote:

Originally Posted by Wackerbarth

It doesn't have to be static when you integrate. Each incremental piece of area is in its own "static" coordinate system.

Ok.
Suppose that constant length of chain is sweeping the grass (n ft).
Let it go from 0 to 360 degrees. The area it sweeps is A1.
Then take straight chain (length 25+n ft) attach one end to the center of the silo. And let it go from 0 to 360 degrees. It's "n ft" part (which is outside the silo) sweeps some area A2.
If you are right then A1=A2 because there is no difference between integrals for n and (n+25) chains.
But A1<>A2

Wacky · 2003-08-22, 12:08

No, you cannot do it that way. You have to look at the area swept by the chain as it moves through an incrementally small change in angle.

In the both cases, the "n ft part" sweeps through an area that looks like a trapezoid. (Arc = chord for very small angles) The triangle that is formed when one end is stationary is a degenerate case.

In both cases, the area is the product of the length (n) and the distance that the midpoint of the chain moves perpendicular to the length (S+n/2).

In my case about wrapping around the silo, the motion on the coordinate system is along the length of the chain rather than perpendicular to it.
So the "far end" of the chain moves n * dT and the near end moves zero, when measured perpendicular to the chain.

c00ler · 2003-08-22, 19:59

Quote:

Originally Posted by Wackerbarth

In the both cases, the "n ft part" sweeps through an area that looks like a trapezoid. (Arc = chord for very small angles)

It depends on the watcher.
If it's static watcher then area looks like a trapezoid (BTW, in your case about wrapping around the silo it looks like a quadrangle)
But if he is attached to the moving polar coordinates system then in all cases area will be like a triangle=circular wedge (Chord=arc for very small angles)

Quote:

Originally Posted by Wackerbarth

In my case about wrapping around the silo, the motion on the coordinate system is along the length of the chain rather than perpendicular to it.
So the "far end" of the chain moves n * dT and the near end moves zero, when measured perpendicular to the chain.

(I must add that we measure these n * dT in polar coordinates.)

The underlined text is what we need.
Because that integral can be used in moving polar coordinates only if "the motion on the coordinate system is along the length of the chain" and one of the ends of the sweeping part of the chain is in the origin of that system.
You hadn't mentioned that condition and your proof wasn't full enough for me.
Now I have no questions to your solution.

rpresser · 2003-09-02, 22:55

... the length of the cow's neck. Or more precisely, the distance between where the chain attaches to the cow and the spot where grass enters the cow's face. It seems to me that if you are giving answers with so many decimal places you should consider how this affects the solution. Not to mention the fact that the chain, having nonzero weight, is going to drag on the ground, so a few inches of chain will be unavailable to the cow....

[remove tongue from cheek now] :D

2003-09-02, 22:55	#22
rpresser Jul 2003 12₈ Posts	But nobody has mentioned ... ... the length of the cow's neck. Or more precisely, the distance between where the chain attaches to the cow and the spot where grass enters the cow's face. It seems to me that if you are giving answers with so many decimal places you should consider how this affects the solution. Not to mention the fact that the chain, having nonzero weight, is going to drag on the ground, so a few inches of chain will be unavailable to the cow.... [remove tongue from cheek now] :D

2003-08-21, 00:14	#12
Maybeso Aug 2002 Portland, OR USA 2×137 Posts	Wackerbarth, Your description was amazingly easy to follow. Congratulations, I'm very impressed. I expected to have trouble following it even with a diagram in front of me. To clear up the very last part, you defined Tf as the angle of the chain when the cow hits the x axis. Is this where your integral stops -- to avoid counting the overlap twice? Consider if the cow continues around. Although the cow and an increasing part of the chain is below the axis, a decreasing amount is above, and sweeps an area of grass that you don't seem to have counted. To clarify: At 3 o'clock the chain is horizontally tangent to the 'top' of the silo, and the cow is at (100 - 50*pi/4, 25). At Tf, the chain has wrapped further around, and the cow is at (x_tf, 0). Now we install a fence along the +x axis (the cow can still go around). We also make the chain retractable. The cow now walks along the fence from x_tf to the silo at 50, the chain retracting to remain straight between the cow and where it is tangent to the silo. Have you counted this area swept by the chain? I thought at first it was the area of your triangle, but you use the triangle to derive Tf. P.S. Your description was so good that I could visualize this possible missing piece without a diagram. ;) [edit] the side of the silo is at 50, not 25. [/edit]

2003-08-21, 02:12	#13
Wacky Jun 2003 The Texas Hill Country 441₁₆ Posts	Thanks for catching my error. You are correct, I forgot that part. All of the above answers are wrong. I left out the area that you indicated. When the cow reaches the x-axis, there is indeed a small portion that the chain has not yet covered. I see two ways to approach it. One way is to shorten the chain while at the same time continuing to wrap farther around the silo. The shortening would be such that the cow travels left on the x-axis until it reaches the silo. The other approach would be to add the area of the triangle that we have omitted and then subtract the piece of the silo that infringes in that triangle. Taking that first approach, the length of the chain along the tangent to reach the x-axis is given by Stan(pi - T) or, in our case -25 tan (T). To integrate that would require integral (tan^2(T) * dT). So let's look at the second approach. When the cow stops at the x-axis, we can form a right triangle. Our chain is the hypotenuse, the x-axis is the base and the height is S * sin (Tf). Counting both this triangle and its image on the other side of the axis, the triangles add 969.7102 sq ft. But this is too much. We must subtract the "lens" which is the part of the triangles inside the silo. 2 * ( pi - Tf ) * 1/2 * 25^2 is the area of the part of the silo between the Tf points and the center of the silo. This is 686.4673 sq ft And finally the triangular part of that wedge gets added back in. 2 * -cos(Tf) * sin(Tf) * 1/2 * 25^2 So the correction is 536.5045 sq ft. And the revised total is 28016.9129 sq ft

2003-08-21, 09:56	#14
c00ler Jul 2003 5² Posts	What about 25516.9262122935 sq ft (Using Wackerbarth's Tf) Or 25516.9260688448 sq ft (Using more precise value of Tf) The formula is (piL^2)/2 + ((L^3)-(L-TfR)^3)/(3R) - (piR^2) Where L - length of the chain R - radius of the silo Tf - the solution of Tf - tan(Tf)=L/R And for short chain(L<=piR/2) formula is (piL^2)/2 + ((L^3)-(L-Tf_2R)^3)/(3R) - (2Tf_2R^2) Where L - length of the chain R - radius of the silo Tf_2 = L/R

2003-08-22, 12:08	#20
Wacky Jun 2003 The Texas Hill Country 10001000001₂ Posts	No, you cannot do it that way. You have to look at the area swept by the chain as it moves through an incrementally small change in angle. In the both cases, the "n ft part" sweeps through an area that looks like a trapezoid. (Arc = chord for very small angles) The triangle that is formed when one end is stationary is a degenerate case. In both cases, the area is the product of the length (n) and the distance that the midpoint of the chain moves perpendicular to the length (S+n/2). In my case about wrapping around the silo, the motion on the coordinate system is along the length of the chain rather than perpendicular to it. So the "far end" of the chain moves n * dT and the near end moves zero, when measured perpendicular to the chain.