The Coolest Commercial, Check it out, it's awsome! Options

[Jeff Krieger]

You couldn't figure out my point, Andy? For the tire I described above, if you position it so that the center of mass makes an angle with the vertical equal to the angle [b] that the ramp makes with the ground and then give the tire a small uphill push, it will "roll" uphill slightly until it reaches the point of (relative) lowest potential energy at a = [pi - b] radians.

And, yeah, it's DOS based software from 1993 and it's pretty basic but it works well and takes the tediousness out of many math operations. If you want to help, I'm willing to take donations that go towards the purchase of Mathematica?

BTW, Andy, why no comment on the emails from the math Profs that I forwarded to you concerning the basic math question that I asked you earlier?

[airsix]

A man does not stand taller standing on someone else.

Everything doesn't have to be a contest of intelect. Can't you accept the fact that YOU ARE BOTH INTELLIGENT and STFU about it?

-Ben M. (still using the old student edition of Mathmatica made for win3.1)

[SirAndy]

i told him to keep it off-list, but he won't ...

[SirAndy]

BTW, Andy, why no comment on the emails from the math Profs that I forwarded to you concerning the basic math question that I asked you earlier?

because i have better things to do than wasting my time with you.
i will reply (offline from this list) when i get a chance.

[Jeff Krieger]

i told him to keep it off-list, but he won't ...

Just go back and re-read this thread, Andy.

[Aaron Cox]

oh great!
the arguments continue!

lets just leave it at..."hey man, what a cool commercial!"

PLEASE (IMG:style_emoticons/default/mellow.gif)

[need4speed]

On the other hand, they probably could have just saved us the physics lesson and done it with computer animation.

[Jeff Krieger]

My previous solution to the "tire rolling uphill" question was incorrect.
Here's the correct (I hope) solution. If you can find a mistake or know of a
simpler solution, let me know.

Let the tire sit just at the bottom of a ramp which has a positive slope. If
the tire's center of mass is a distance r from the tire's center and the
center of mass is located (initially) at the origin of the x-y coordinate plane (see my
diagram below), it's height (y) above ground is given by y = r(a-sina)sinb +
r(1-cosa)cosb. Here "a" is the angle formed between the radius of the center
of mass and the angle -pi/2 + b. "b" is the angle that the ramp makes
with the horizontal axis.

Since the tire's mass (m) and the acceleration due to gravity (g) are both
constant, the critical points of the tire's potential energy (mgh) equation
are determined completely by the the height of the tire's center of mass
above the ground h = y = r(a-sina)sinb + r(1-cosa)cosb. For fixed r and b,
dy/da = r(sinb)-r(sinb)(cosa) + r(cosb)(sina).

For the specific case where r = 4 and b = pi/6, dy/da = 2sqrt(3)(sina) -
2(cosa) + 2 which has critical numbers a = 0 + 2n(pi) and a = 4/3(pi) +
2n(pi) where n is an integer. a = 0 corresponds to the tire's center of mass
being at the origin and having 0 potential energy, a = 4/3(pi) corresponds
to the tire's center of mass being pi/6 radians after TDC (see my diagram
below) and is a local maximum. If at this point the tire is given a slight
uphill push, then it will roll uphill until it reaches a = 2(pi) radians
which is a local minimum for the potential energy equation. The rotation of the tire from a = 4/3(pi) to a =
2(pi) radians moves the center of the tire 4(2 - 4/3)pi units up the hill.

(IMG:http://persweb.direct.ca/aschwenk/diagram.jpg)

Here is a graph for y = r(a-sina)sinb +
r(1-cosa)cosb for the specific case when r
= 4 and b = pi/6. (In this graph x = a).

(IMG:http://persweb.direct.ca/aschwenk/PE2.jpg)