| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1734.1 | | CSC32::CONLON | | Fri Mar 26 1993 10:38 | 9 |
| The problem comes from my (refresher) course in Calculus II (which
I'm taking to prepare myself for a few upper division math classes
I need to take later this year.) It's not a homework or home test
problem, by the way. We were just asked to give it some thought.
We'll be looking at this problem over the next couple of weeks (the
instructor does have the solution which I will post when he gives
it to us.) I've been thinking about it and wondered if anyone here
might have some ideas.
|
| 1734.2 | | STAR::ABBASI | i am therfore i think | Fri Mar 26 1993 12:10 | 14 |
| how about:
| + +
| / \ / \
| / \ / \ --->length = 1 feet
|/ \ / \
+---------+---------+----------------------------------------->x
0 h 2h
feets
____
where h = \/ 2
\nasser
|
| 1734.3 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Fri Mar 26 1993 17:20 | 9 |
|
Well, from a *height* perspective, it might not be a bumpy ride.
But, from a *power* point of view, I suspect it will be quite bumpy. The lurch
will be felt as a fore-aft jostling instead of an up-down jostling, right ?
/Eric
|
| 1734.4 | | CSC32::CONLON | | Tue Mar 30 1993 15:30 | 22 |
| No answer yet, but more info from the instructor: [Pls excuse the
drawing - not to scale.] Only 1/8 of the square is being considered
here (the triangle ABC formed from the axle to the bottom right corner
of the square.)
....................
. .
. . ___
2 . A .--------------- Axle is \/ 2 above
. | . . ground (x-axis) at all
. | . . times
D........_C_.......B
/ | \ When square rotates, the
/ | \ corner B touches x-axis
________________/_____|_____\_____________
*******
Length of 1/2 the curve = 1.
Bottom of square DB is tangent to the curve (and he did say it was
curved). As the square rotates, the height of point A is constant.
He said to consider the angles which lines AB and AC make with the
perpendicular from A to the x-axis.
|
| 1734.5 | This should reduce the search space :-) | VMSDEV::HALLYB | Fish have no concept of fire. | Wed Mar 31 1993 09:54 | 5 |
| > Bottom of square DB is tangent to the curve (and he did say it was
Ahh -- so the tangent to the curve is a straight line!
John
|
| 1734.6 | another viewpoint | CGDEIS::GERTLER | | Mon Apr 12 1993 23:52 | 11 |
| another way to view this problem is to place the square on a flat
surface and describe the curve the center of the square makes as
the square rotates along the ground. This should lead you to:
sqr(2) * sin(x+45)
as x goes from 0 (square lying flat) to 45 (square on vertex). To
get the road curve, you'll just need to slide this 45 degrees out
of phase.
DAG
|
| 1734.7 | The "forefactor" | VMSDEV::HALLYB | Fish have no concept of fire | Tue Apr 13 1993 10:01 | 9 |
| Circular functions have a period of 360 degrees, yet this problem
appears to have a period of 90 degrees -- one rotation and you're back
to the original position. So any solution is likely to take on the
form
sin(4*theta+K)
With other ingredients such as absolute value, sqrt(2) multiplier, etc.
John
|
| 1734.8 | | CSC32::CONLON | | Tue Apr 13 1993 11:28 | 8 |
| Well, we got the solution from the instructor.
The curve is an upside-down catenary ('hanging cable'). So it turns
out that if you take the shape of a hanging cable and turn it upside
down, you can roll a square over it.
__
y = \/2 - cosh (x)
|
| 1734.9 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Tue Apr 13 1993 11:43 | 81 |
| I can only derive the upside-down catenary solution by
assuming that the vertical line from the center of the square
to the x-axis passes through the point of tangency between the
square and the curve.
Let o's represent the curve and *'s represent the square and
make the assumption mentioned in the previous paragraph:
*
*
* *
*
. *
* |
| *
* | *
*o o
* o| o
. o | o
. o | o
--.------o-------------o----------
_
Given that assumption, the constant height \/2 from the
center of the square to the x-axis is a sum of two parts,
the function value y and the diagonal of a right triangle
within the square with one leg having length 1 (center
of square to side of square) and one leg lying along the
side of the square tangent to the curve. The length of
that diagonal will be the secant of the angle a at the
center of the square. Putting this together gives:
_
y + sec a = \/2
That angle a is also the angle the side of the square
makes with the x-axis (see the dots extending the *'s
of the square). The tangent of that angle is the slope
of that side of the square and also is the slope of the
curve at the point of tangency:
y' = tan a
_
This can be solved by substituting z = \/2 - y, z' = -y'
sec a = z
tan a = -z'
2 2
and using the identity sec a = 1 + tan a, so that
2
z' = - sqrt(z - 1)
The solution to that is
2
dz/dx = - sqrt(z - 1)
2
dz / sqrt(z - 1) = - dx
-1
cosh z = - x + C
z = cosh(- x + C) = cosh(x - C) (cosh is an even function)
Therefore
_
y = \/2 - cosh(x - C)
If you let the curve peak at x=0, then C = 0, and the
solution assuming the vertical line from the center of
the square to the x-axis passes through the point of
tangency of the square and the curve is
_
y = \/2 - cosh x
_
for x such that cosh x <= \/2; make the function periodic
by repeating that section of the curve.
Dan
|
| 1734.10 | | CSC32::CONLON | | Tue Apr 13 1993 11:49 | 10 |
| RE: .9 Dan
> I can only derive the upside-down catenary solution by
> assuming that the vertical line from the center of the square
> to the x-axis passes through the point of tangency between the
> square and the curve.
Yes, the instructor did say that the solution is based on this
assumption.
|
| 1734.11 | Other solutions? | CADSYS::COOPER | Topher Cooper | Tue Apr 13 1993 15:28 | 5 |
| In other words -- it is *a* solution but not necessarily the only
solution. Other solutions might be found if that assumption is
relaxed. So we are not done yet.
Topher
|
| 1734.12 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Wed Apr 14 1993 12:09 | 5 |
| > So we are not done yet.
That goes without saying. :-)
Dan
|
| 1734.13 | | CSC32::CONLON | | Wed Apr 14 1993 12:12 | 11 |
| RE: .11 Topher
> In other words -- it is *a* solution but not necessarily the only
> solution. Other solutions might be found if that assumption is
> relaxed.
Absolutely! Also, if the assumption is false, then it isn't even
necessarily *a* solution. :>
(It is, however, the only solution my instructor has offered.)
|
| 1734.14 | | CSC32::CONLON | | Wed Apr 14 1993 12:12 | 3 |
|
Thanks, Dan (and everyone else who replied)!!
|
| 1734.15 | Edge-ing them on | VMSDEV::HALLYB | Fish have no concept of fire | Wed Apr 14 1993 14:33 | 31 |
| The boundary conditions may provide some help as well.
o o o o
o o o o
o o o o
o o o o
---------o-------------o-------------o-
O A B
At points A and B (and so on) one needs a "smooth" transition from
one segment of the curve to the next. I take this to mean that at
point A the curve must form a 90 degree angle, as measured by a
limiting process. If the angle is \less/ than 90 degrees then the
square doesn't hit the x-axis, and therefore the center of the square
would be more than SQRT(2) above the X-axis. If the angle is \more/
than 90 degrees then there is an interval (not a point) in time when
the only contact with the ground is the corner of the wheel, meaning
the center of the wheel experiences a bump, which is not allowed.
You need a clean "fit" into the groove at A, B, ...
When I solve for point A I get approximately .8813736 and checking
the derivative we see that the slope of the tangent is indeed -1 from
the left and +1 from the right. So the proposed solution fits the
boundary criteria, and the idea of just repeating the function in
segments appears to work out OK.
Maybe this can be used to show the tangency assumption is the only
valid approach. Then we'd be done.
John
|