Last week I found my notebook re. a game I was developing a year ago. 
I was, and still am, stumped:  How much work is required to tip over a
chariot?

For simplicity, let's imagine a box: 70 kg mass, 1.8 m high and 1.4 m
wide.  It is going around a curve, so there is a centripetal force
acting on it, parallel to the ground.  Let's also ignore friction.

After reviewing some calculus and definitions of torque I tried out
some equations and came up with an equation which I don't know how to
write in this text editor.  Using the dimensions mentioned above I
found an answer of 1252 Joules, but I could be completely wrong.

I'm looking for the work required to tip the object to its point of
equilibrium, since beyond that point gravity would pull it down onto
its side.

Thank you!
Todd H.

Hello tharrop-ga 

  Since you left the question 'on' 
  I suppose you want still another answer.
 
  Work required to tip the object is the difference between the
 energy of the object 'standing on its base' , which is
  E0= m *g * .9    
 and energy of the object 'standing on edge'
  E1= m * g *  1.14017543

 Difference is 0.240175425 * 9.81 * 70 = 164.928464 Joules.

  This is the Bozo's result but with more accurate numbers.

 m is mass = 70 kg  and energy is m * g * h ,
 where h is height, distance between center of gravity and some arbitrary
 reference level (surface of the Earth).

g is 9.81 on the surface of the planet,
http://physics.webplasma.com/physics07.html
 
 The height for the edge is sqrt((.9 * .9) + (.7 * .7)) = 1.14017543
 The height for the base is  1.8/2 = .9
Comment on comments:

tibol computes Moment, which is NOT work you are looking for.
      Besides, statement "1 kgm = 9.8 Newtons" is just plain wrong.
               Probably he means 1 kg weights 9.8 Newtons,
               which is true, but it is NOT an '=' sign.

 myoarin-ga  interprets the question differently.
             The energy needed to tip object is independent of the 
             mechanism by which tipping is accomplished.
             Just in  case you are interested, force on a curve is
             discissed here:     

http://answers.google.com/answers/threadview?id=509922

Above we assume that (right bottom) edge of the box is fixed, perhaps
by some boulder or ridge (so we can tip in absence of the friction).

More on tipping is shown here:
http://www.unc.edu/depts/appl_sci/ortho/biomechanical/tipping.html

---

Clarification of Answer by hedgie-ga on 14 May 2005 05:05 PDT

Well, Tharrop

 This perhaps was one of the longest debates resulting from a
technical question. Unfortuntely, most of it was unrelated and part of
it was incoherent, but there were some valuable insights.

 I think we have come to an end, (I mean natural death of the debate)
 and so I will comment on the broader question: 

          when does a vehicle rolls-over? 

One quantity which can be calculated is 'rollover velocity' - which assumes 
circular track of radius and SLOWLY increasing speed v.

Actually, this 'critical velocity' was calculated in the comment:
--------------------------------\\
Subject: Re: Work required to tip over an object 
 From: snesprogrammer-ga on 13 May 2005 01:47 PDT

The chariot will tip when mv^2h/r > mgw

------------------------------------------//

 However, after  one edge lifts of the track, things get complex:
  
   Since the 'righting torque' will keep decreasing to zero as roll-angle
   is increasing, vehicle will roll over UNLESS driver reacts: 
   unless s/he decreases the speed or steer to right the roll -
   which every driver will try to do.

  We know that stunt drivers can ride on two wheels, indefinetely, by 
   adjusting the velocity and direction -- a bit like we all do on a
bicycle.

This makes the whole problem a complex problem from the Control Theory:
 It does not have an answer given by a single number. Rather, one has to 
keep solving the differential equations which shows how the
'generalised coordiantes' of the car change with time in response to
controls
 (brakes, gas, steering wheel .. which would have to be specified and described).

 Such simulations are done for 'real car' design and simplified for games,
 but it would  exceed the scope of this question, and probably your
patience as well to dive into that.

Here, just for illustration, is one technical  paper, which gives the
equations for one such control problem
[
http://www.eng.auburn.edu/~dmbevly/gavlab/pub-pre/Using Scaled
Vehicles - Rollover Prop.pdf
]  whole  [ ..] is the URL - it may need to be pasted into a browser.

Basically, equations need to solve the combined dynamics :
 car moving forward,
and turning in response to steering 
and the roll-dynamics, where Torque (weight + inertia induced)
acts on the Moment of inertia - all that around an axis,
which is changing direction with time.

One can leave the control to the player of the  game,
 or model the driver as well.

There are technical studies and simulations done for the industry needs,
which are probably too complex for what you may actually need.

Nevethless, here are few - they may be useful to illustrate the
equations and their complexity and look of the results:


2- Yaw/Roll model

 

The UMTRI Yaw/Roll Model was developed for the purpose of predicting
the directional and roll response of single and multiple articulated
vehicles engaged in steering manoeuvres, which approach the rollover
condition. It should be noted that the model does not permit the
simulation of braking manoeuvres. However, it does permit the analysis
of unconventional heavy vehicles, such as straight trucks (up to 4
axles), tractor/semitrailer ((up to 8 axles), doubles (up to 11-axle)
and triples (up to 11-axles).


4- Static Rollover Simulation Model

 

This model was developed by UMTRI and able to calculate the rollover
threshold of articulated vehicles during steady state turning
manoeuvres. This model incorporates the effects of the torsion
compliance that exists in the tractor frame, fifth wheel and trailer
structure compliance. This model is very sophisticated and proven to
be accurate in comparison with the tilt table tests and yaw/roll model
predictions of the static rollover threshold.

http://www.vss.psu.edu/VSRC/vsrc_simulation_lab.htm


The Simple Unstable Vehicle: a manual control task

http://www.simongrant.org/pubs/thesis/4/0.html

Balancing a pole  or an inverted pendulum
http://www.cs.colostate.edu/~anderson/code/pole-tcl.html
 



he HD2002 stunt car simulation project
http://members.xoom.virgilio.it/Devas/

Good luck with your project

Hedgie

There was an abundance of scientific, pedantic and philosophic debate,
so it was impossible to know what was the answer.  I only wish my
question had been better.  There was an incredible display of
knowledge, however, and I'm very impressed at the quality of research.
 Thanks!

Work is force multiplied by distance MOVED IN THE DIRECTION OF THE
FORCE (i.e. a scalar produced by a scalar product of 2 vectors).  
Torque is a vector product of force and a different distance.

I should think the distribution of mass in the chariot as well as
details of the cornering will be necessary information.  Neglecting
these details I think you need to lift the centre of mass from 0.9m to
1.1m (arrived at using a right-angled triangle).  That's 0.2 * 70 * 10
or around 140 J.

Your assumption is too simple for the actual condition of tipping over
a chariot that is running around a circular or curved path.

Anyway let us calculate it based on your assumption.
But we cannot ignore the friction. Maybe you mean that the friction is
good enough so that the box will not slide radially or outward away
from the center of the curved path.

I assume the box is homogenous, so its centroid is at the exact center
of the box. The centroid is 1.8m/2 = 0.9m above the ground, and 1.4m/2
= 0.7m from the edges of the width of the box.

Whether the box is moving or not, to tip it over about the outer
edge---the edge farther from the center of the curved path---of the
width, the weight of the box will tend to counteract the tipping. This
is the righting moment of the box.
Moment, like work, is force times perpendicular distance.
For the righting moment, Force is due to the weight of the box acting
on the centroid. Since 1 kgm = 9.8 Newtons, then this force is (70
kgm)*(9.8 Newtons / 1 kgm) = 686 Newtons.
Then, the righting moment is (686 N)*(0.7m) = 480 Newtom-meter, or 480 joules.

And that is the work you are looking for.
It is not 1252 joules.

That tips over the "box", whichever of you is correct, but tharrop is
asking about a chariot going around a curve, so it is going to be a
problem of the speed and the radius necessary to generate tilt past
the righting moment, the final answer being the work required to move
the chariot at the velocity necessary on a curve of a given radius to
make it tip over.  Bozo99 is right, the radius of the curve must be
defined.

At least I think so...

Hedgie,
You'r just running circles around me, dizzying!  :)

Whoa, hedgie-ga, you messed up. Big time.
Did you know what you talked about in the answer you posted for this question?

" The height for the edge is sqrt((.9 * .9) + (.7 * .7)) = 1.14017543
The height for the base is  1.8/2 = .9"

Yeah?

Hedgie-ga, the box is on the ground.
So the heights of the edge and base are both zero.

---------
If you interpeted the question as if the box is raised above the
ground at some arbitrary height, then the box would not tip at the
edge. The box would tip over the outer wheels, or the wheels farther
from the center of the curved path. We don't know the width of the
wheelbase, you did not give assumption, so we could not imagine the
distance of the fulcrum/pivot where the box would tip.
[If the width of the wheelbase were 1.4m also, then the horizontal
distance between the centroid of the box and the tipping pivot is 0.7m
also.]

In this interpretation, or in the above interpretation, or in any
interpretation, the height of the centroid from the edge is not your
sqrt[(0.9)^2 +(0.7)^2]. That is only the distance of the centroid from
the edge.

Your
"energy of the object 'standing on edge'
  E1= m * g *  1.14017543"
is way, way off.

The force due to the 70kgm is pointing vertically down. The 1.14017543
m is oblique to the direction of the vertical force.
So how can the energy on the edge be the vertical force times the oblique distance?

--------
Your answer is full of holes. The "holes" mentioned above are only some of them.

You sure you believed your answer?

--------
"tibol computes Moment, which is NOT work you are looking for.
      Besides, statement "1 kgm = 9.8 Newtons" is just plain wrong.
      Probably he means 1 kg weights 9.8 Newtons, which is true, but
it is NOT  an '=' sign."

So you did not undertand my comment.
If you were sharp enough, then you could have understood that
overcoming the righting moment of 480 jopules is the work needed. The
value/measure of the work is 480 joules.
[Actually, the work to tip the box should be a tad more than 480
joules, because at 480 joules, the box would be in equilibrum only.]

1 kgm weighs 9.8 Newtons?
Zeez, where did you get that?
Have you read/heard/seen somewhere that Newton is a unit for weights?

Weights are in pounds, kilograms, stones,.., and their denominations.
But never in Newtons.

1 kgm = 9.8 Newtons is shown everywhere. See you books, the internet,...

ticbol-ga 

      I am really sorry if I ruffled your feathers by my comment;
 I do not argue with comments, usually.
 What got me this time was your use of 'kgm'.

 I am using SI units, described e.g. here:
http://physics.nist.gov/cuu/Units/

So I did not say
1 kgm weighs 9.8 Newtons?


I said
 1 kg weighs 9.8 Newtons

To wit:
kg is unit od mass.
weight is a force, by which a mass is attracted to the Earth

One reason for using SI units (kg for mass, N for force) is that
the relationsgip is less confusing, as explained here
http://www.npl.co.uk/mass/faqs/massweight.html

If you try SI units, you will like them too, I bet.
So, forgive me and do not take it personally. It is not
your fault. It was Jefferson ho got US to adopt the British
units - and they still teach them in schools - 
to detriment of US students.
Hedgie

Ticbol, hedgie's right on this one, both about the answer and about
Newtons being a unit of weight.  The energy required is just the
difference in potential energy between the box on its side and the box
balanced on its corner.  Weight is just the force on an object due to
a gravitational field, so any unit of force can be a unit of weight. 
I agree with you about systems of units though--who cares?  I use SI
units because I got used to them in school, but it really doesn't
matter much.

I thought I posted this comment this morning. Where is it it now?

Let me post it again.

---------
To racecar-ga.

No. You and hedgie-ga are wrong on the answer and on the Newton as a
unit of weight.

On the Newton, the reason is no-brainer. No need to think why Newton
is not used as a unit for weights.
Nowhere can you find weights expressed in Newtons.

For the answer, the difference in energies [in the box from at rest on
its base to its position on its edge] is not the work needed to tip
the box about its edge.
Have you thought about the energy needed to put the box from rest on
its base to its position on its edge?
That is the work needed.

Review your Mechanics.

---------------
More or less, that was what I thought I posted about 11 hours ago
before I went to work this morning.
I was hoping you and/or hedgie-ga would say something about that, so
that I could correct you again.
So, I will leave this for a while. If you want to comment, racecar-ga,
let us have it.

A newton is, in fact, the metric unit of weight.  Weight is simply the
force of gravity exerted on a object of a particular mass.  The metric
unit of mass is the gram, and the metric unit of force (and therefore
weight) is the newton.  The SI unit of mass is the slug, and the SI
unit of force (and therefore weight) is the pound.

One slug is 14.5939 kilograms, incidently.

linezolid-ga 
               Did you say:

 " The SI unit of mass is the slug"
       
  Surely you are joking. There are many metric systems (cgs, MKS ..)
  and one of them, SI, was universally accepted as THE system,
  recommended for all measurements.

I have posted a link to it above; Here is another link:
http://www.unc.edu/~rowlett/units/sipm.html

it says:
"The key agreement is the Treaty of the Meter (Convention du Mètre),
signed in Paris on May 20, 1875. 48 nations have now signed this
treaty, including all the major industrialized countries. The United
States is a charter member of this metric club, having signed the
original document back in 1875..."

Hi Ticbol,

Hedgie already posted a link showing that the newton is a unit of
weight, so I'm not sure why you're still arguing.  Here's the link:

http://www.npl.co.uk/mass/faqs/massweight.html

Here are 5 more (of 1000's I could have chosen) which refer to weights in newtons:

http://hypertextbook.com/facts/2004/WaiWingLeung.shtml
http://hyperphysics.phy-astr.gsu.edu/hbase/mass.html
http://www.answers.com/topic/newton
http://theory.uwinnipeg.ca/mod_tech/node25.html
http://www.grc.nasa.gov/WWW/K-12/BGP/Sheri_Z/thrustwtrat_act.htm


ticbol wrote:
For the answer, the difference in energies [in the box from at rest on
its base to its position on its edge] is not the work needed to tip
the box about its edge.

YES IT IS.

ticbol wrote:
Have you thought about the energy needed to put the box from rest on
its base to its position on its edge?
That is the work needed.

THAT IS THE DIFFERENCE IN ENERGY BETWEEN THE TWO STATES.

ticbol wrote:
Review your Mechanics.

REVIEW YOURS.  I TEACH IT, WHICH TENDS TO KEEP IT FRESH IN MY MIND.


Your earlier comment did get posted--I saw it.  It has since
disappeared, probably because the google editors did not like its
pejorative style.

Good, racecar-ga. Now you responded.

----------
On the Newtons as unit of weight, I still don't see anywhere where
anything is said to weigh some Newtons.
The weights are either in pounds or kilograms (or in their
denominations) or stones, etc. No newtons.

The sites you mentioned---I went to one---can say that Newton is the
actual unit of weight, that many items in groceries are in the
one-Newton range of weight. But, still, these items are in pounds or
kilograms in their labels.

-------
On the "difference in energies...as the work needed..."

See this:

a) Box at rest on its base.
Potential energy, based on the base or ground, is (70kgm *9.8)(0.9m).

b) Box balanced on its edge.
Potential energy, based on the ground, is (70*9.8)[sqrt(0.9)^2 +(0.7)^2]

c) Difference in potential energies between the two positions of the
centroid is (1.14 -0.9)(70 *9.8) = 165 joules.

That is your, and hedgie-ga's, answer.

Now consider these:

d) if the box were just lifted vertically so that the centroid is
raised from the initial 0.9m height to 1.14m height, would the
difference in potential energies be not 165 joules also?
So the work needed to lift the box vertically only is already 165 joules.

e)But the centroid when the box is balanced on its edge is 0.7m
horizontally away from the vertical 1.14m height.
The new height is still 1.14m, okay, so you (and hedgie-ga) are saying
the work needed to place the centroid 0.7m away is still 165 joules?

Exploding that, you are implying that as long as the difference in
height between the initial position (0.9m high) and the final position
(1.14m high) of the centroid, the work needed to transfer the location
of the box will always be 165 joules? That if the box were lifted,
say, 1 km away, only 165 joules is necessary to accomplish it?

I don't need to ask that in capital letters, racecar-ga.

--------
So you teach Mechanics?

Are you sure?

Besides the questions above, in your version of Mechanics,
moments---turning moments, righting moments---are not included?

Summation of clockwise moments is equal to the summation of
counterclockwise moments in a body in equilibrium, you know....

-------
Oh, so the moderators could have erased my missing comment?

Umm....

I hope they let this comment be posted. I hope they understand that
arguments or arguings are not always done in "Excuse me, Sir, I think
you are wrong in your comment..." styles.

I argue, I attack. My style.

Belittle? No.

Aggressive? Yes.

Nothing personal.

I used capital letters to make it clear which were my words and which were yours.  

I am not going to argue any more after this about newtons.  The
discussion started when, in an earlier comment, you wrote:

1 kgm weighs 9.8 Newtons?
Zeez, where did you get that?
Have you read/heard/seen somewhere that Newton is a unit for weights?

As we all know, and you have now implicitly conceded, the answer to
your question is that yes, a newton is a unit of weight.  Who cares
whether it's the most popular one with the people who print the labels
for packages of cookies?  This is a physics question, posted in the
science category, not a cullinary merchandise question.

What is more important is your misconception about the conservation of
energy.  In the absence of friction, the work required is exactly the
difference between the total initial energy and the total final
energy.  If we take a box that is initially at rest, and raise its
center of mass by a certain amount, ending once again at rest, that
increases the gravitational potential energy of the box, and the
amount of that increase is the amount of work required.  As long as
there is no kinetic energy in the initial or final states, and no
friction during the movement, this is true.  It matters not at all
that the box happens to be in a different orientation or that it is
horizontally displaced.  This may be a difficult concept to get used
to, because you know that in the real world, to move a box across the
room takes some work.  But that is only because of friction.  The
theoretical answer to your question about lifting something .7 m up
and 1 km sideways is that yes, it requires the same amount of work. 
That is, if you had a frictionless ramp 1 km long and .7 m high, it
would take the same amount of work to push the box up it as it would
to lift it straight up by .7 m.  It's just that you apply a much
weaker force over a much longer distance.  That is what makes ramps
useful--you can lift things up with less force.  Even in the real
world, the friction acting when you tip a box up on one of its corners
is pretty much negligible, so hedgie's answer is basically right not
just in an 'ideal physics world' but also in the complicated messy one
we live in.  You also seem concerned about the rotational energy the
box has while it's being lifted.  Yes, rotating bodies have kinitic
energy (it is .5 I w^2, where I is the moment of inertia and w is the
angular velocity).  But the beauty of conservation of energy is that
you don't have to worry about exactly where all the energy is at every
moment in time.  If you know the initial energy and final energy,
which is simple in this case because both are pure gravitational
potential energy, the work is just the difference between them.

To answer your last questions, yes I really teach mechanics.  And I
really have a degree in physics.  Instead of treating this as a
debate, stubbornly sticking to your position and looking for ways to
support it, you might consider the possiblity of learning something
from this exchange.

No, I did not, will not, concede, implicitly or explicitly, that
Newton is a unit of weight. Newton is a unit of force.

As to your long explanation re conservation of energy anf or work,
Zeez, you sure believe what you have just explained?

I cannot believe a guy who teaches Mechanics, never mind a graduate or
degree-holder in Physics, do understand conservation of energy, and
work, the way you posted it here. Unbelievable.

Think it over, racecar-ga. 

I am embarrassed myself that I am arguing with you this
"misconception" on my part. On my part. Whoa!

Back off for a minute and look at this 70 kg 1.8 x 1.4 m block.

For one thing, we are assuming 100% friction on the ground; we are
assuming that it is going to be be tipped as though it were hinged to
the ground.  Right? 1.

And then, as we all know from experience with moving tall items of furniture,
when you start to tilt them, initially you have to raise half the
weight - the other half is resting on the ground.  Right? 2.

But immediately gravity starts to draw on that portion of the item
(the block) which is tilted past the vertical line of the "hinged"
side.  Right? 3.

This reduces the work necessary to continue to tilt the block as more
and more of its mass moves past this vertical line.  Right? 4.

And then  - whoops! -  the block starts to tilt by itself, before we
expected it to, because as the weight decreased we didn't reduce our
effort accordingly, increasing the speed  - momentum -  ...  Right? 
5.
But we are ignoring momentum it seems, so this is all happening in
super slow-motion - no wasted effort of accelerating the tilting
motion.   Right? 6.

(Of course, if we were very skilled at this, we could apply just
enough effort to accelerate it and let its momentum carry it to the
critical angle, slowing down then to a stop, and we would have
expended a greater amount of effort in a shorter period of time, but
the "work" would have been the same (disregarding the fact that
friction with the atmosphere is theoretically involved and different,
being a function of velocity, but not at the speed we tilt this
thing).

So, at no time do we have to raise 70 kg, and only for an instant at
the start do we have to raise 35 kg.

Is this reflected in your calculations?  

Myoarin

Yes, myoarin.  It is possible to get the answer by calculating the
force required to lift one corner of the box at every instant and to
integrate that force multiplied by the distance moved. The force, as
you point out, starts out at 35 kg * 9.8 N/kg and decreases after
that.  But the height to which you have to raise the corner to reach
equilibrium is 0.8595 meters.  In the end you would get the same
answer as you get by just multiplying the mass by g and 0.2402 meters
(which is the height you must raise the center of mass) but with a lot
more effort.   That is what I meant earlier about the beauty of
conservation of energy.  You don't have to worry about the
instantaneous force, power etc.  If you know the initial energy and
final energy, the energy required to bring about the change is just
the difference.

Ticbol--so you can acknowledge that thousands of websites state that
the newton is a unit of weight, but you won't concede that it's true. 
You can scoff at my explanation without even attempting to find a flaw
in it.  You are illogical, so it is pointless to argue with you.  This
is my final attempt.  Instead of showing why hedgie's way is right,
I'll show you why yours is wrong.  You wrote:

>Moment, like work, is force times perpendicular distance.

Yes, moment (in physics we call it torque) is force times
perpendicular distance.  But when we calculate work by multiplying
force by distance, it is the parallel distance (the distance moved in
the direction of the force).  So while the units are the same
(newton-meters for example), torque and work are not at all the same
thing.

>For the righting moment, Force is due to the weight of the box acting
>on the centroid. Since 1 kgm = 9.8 Newtons, then this force is (70
>kgm)*(9.8 Newtons / 1 kgm) = 686 Newtons.
>Then, the righting moment is (686 N)*(0.7m) = 480 Newtom-meter, or 480 joules.

You have correctly calculated the torque required to begin to rotate
the box.  But you have incorrectly stated that the torque (moment) is
the same as the work. Work is torque times angular displacement.  The
angular displacement of the box when it's at its equilibrium point is
atan(1.4/1.8) = 0.661 (radians).  If the torque stayed constant at 480
Nm, the work would be 480 Nm * 0.661 = 317 J.  As soon as the box
begins to rotate, however, that torque begins to decrease, going to
zero as you approach the equilibrium point.  So to get the work that
way, you'd have to do an integral over the angular displacement of the
instantaneous torque.  In which case you'd get the same answer hedgie
does.

1 is right.

2 is wrong. 
Once the box is lifted, if the box is rigid, the whole weight is off the ground.

3 is wrong.
At any time, whether the bx is resting on the ground or it is being
lifted, the gravity pulls on the whole mass of the box---not only on
the portion of the box past the pivot.

4 is correct.

5 is wrong.
The box will not tilt by itself yet. The centroid has not passed the
vertical line of the hinge/pivot. If you let go of the box this time,
the box will fall down to its original position.
So "work" is still needed until the centroid reaches the vertical line
of the pivot.

6 .... ?
We don't ignore the momentum. If the box is moving, there will always
be momentum. Depending on the force you apply, there will be
deceleration, constant speed, or acceleration in the movement on the
box. F = m*a. The "m" is contant, so the "a" depends on the F.

The rest below the 6 is wrong.

---------------------
The box sits on its base on the ground. Gravity pulls down the whole
mass of the box. The resultant of this pull can be assumed to be
acting as a concentrated force acting on the centroid of the box. To
lift the box at the edge opposite the pivot, effort is needed to raise
that edge, and hence the whole box, off the ground, even for just
0.0001 centimeter high.
What is preventing your effort to initially lift the edge? It is the
mass, hence the weight, of the box.
Imagine a lever, 1.4m long, resting on the ground that is hinged at
one end. The whole weight of the box is "concentrated" at 0.7m from
the hinge. Your hands are at the other end, or 1.4m from the hinge.
Now, how much force do you need to lift the weight of the box even
just 0.000000001 centimeter off the ground? Still remember how to
solve this simple problem?
In your solution you unknowingly are applying the rule that
force*distance from the hinge is equal for those going to the right of
the hinge and those going to the left of the hinge. These
forces*distances are "moments", "work", etc.

racer-ga, I have to go to work. I will answer your latest
comment--which I have just seen now---in about 12 to 14 hours from
now.

Hedgie perfectly answers your question.  I just wanted to comment that
the spinning wheels on a chariot would resist tipping somewhat, so
some energy would also be required there.  It would have the effect of
slowing the wheels down if I'm not mistaken.

You might want to calculate the initial rotational energy in the
wheels and decide whether or not it's negligible.  Of course, reducing
the chariot to a box of uniform density is probably a bigger deal, and
in a gaming context you're probably right to simplify.

Okay, racecar-ga, I am back.

1.)Re Newton is not a unit for weight.

No, I don't concede that Newton is a unit of weight even if websites says so.
Again, if Newton is the unit of weight in metric system, then why is
that not adapted in real world? You buy 20 Newtons of apples? In your
personal data, in applications, etc, do you write your weight as, say,
833 Newtons instead of 85 kgm?

As I have said, the website that I visited implied that Newton is the
unit of weight. I don't believe that.

2.) Re the answer to the question above.

So, as you said, work is force times distance traveled parallel to the
direction of the force. [As opposite to moment is force times
perpendicular distance.]. Okay, let me follow that.
The chariot/box is rounding the curve path. Centripetal force tries to
pull the chariot/box toward the center of the curved path. Physics
says an equal centrifugal force counteract this centripetal force.
This centrifugal force, F = m(v^2)/R, is horizontal. It acts
horizontally at the centroid of the chariot/box. The horizontal
distance that the centroid will travel from standing at base to
standing on edge for the chariot/box is 0.7m.

Am I correct if I say "work" is done by the centrifugal force, which
is horizontal, times 0.7m, which is horizontal too?

For the vertical component of the work needed, as you and hedgie-ga
computed, is (70*9.8)(1.14 -0.9) = 165 joules.

I say the work needed is 480 joules. So 480 -165 = 315 joules, and
that is the work done by the centrifugal force.

3.)I opened an old Physics book to review this work = force times
parallel distance. Why you insist that work depends only on the
difference in heights between two positions. That horizontal distance
does not count.
And there in the book, it says, work depends only on difference in
heights forconservative forces like that caused by gravity.

And for non-conservative forces, work is not depended on difference in
heights. Voila!
So it is also written in books what I believe re work.

This problem above is not about work due to gravity only. It is not
about connservative forces only.
The friction, which prevents the chariot/box to slide outward
radially, provides the non-conservative force. And, the fixed location
of the centroid from the edge/pivot contributes in a way to the effect
of this force.

4.)...which lead me to come back to Summation of moments about a point
is zero for an object in equilibrum.

If an energy/moment/torque/work/(force*distance) is to be overcome, an
equal energy/..... is needed to start the motion. A little more will
start the motion.

Can you not see conservation of energy there?

Ticbol asks:

"why is that not adapted in real world? You buy 20 Newtons of apples?"

 Of course not. That's the point Ticbol: I buy 2 kg of apples.

 (The symbol kgm for kg is REALLY obsolete, could you conceed that
  and use the SI symbol for unit of mass?)

 That's because when we shop for groceries, we care about the mass, not weight.
To wit:
 On the Moon, I would still aks for 2kg of apples and get the same ammount,
 but they would weigh less. Do you  admit that weight is a force?
 
On the Moon things weigh less - yes or no?

 In some countries in Europe they are introducing in science
 classes new verb 'to mass': 

 Like " These  apples mass 2kg" 

to replace older  " These  apples weigh 2kg"

 since the older way is confusing to some people.
   
How would you like that convention?

I also have a degree in physics, but unlike racecar-ga I haven't
taught this in awhile (so if I goof up, definitely defer to
racecar-ga).  Instead of the same thing being said again and again,
let me try to explain it in a new way:

-------------

Is a kilogram a measure of weight?
No.  It is a measure of mass.
Take a kilogram here on earth and it will weigh about 2.2 pounds. 
Take a kilogram up to the moon (where the gravity is roughly 1/6 of
earth's) and it will weigh about 0.37 pounds.  But it's mass is still
one kilogram.  Free floating in space: weight=0 pounds, mass is still
1 kilogram.

Those measurements of grams or kilograms, etc on grocery items are
referring to the mass.

The problem is, since everyday life is here on earth, people
colloquially refer to kilograms as a measure of weight (in other
words, _implying_ that the gravitational force used is Earth's
gravitional force).  Heck, even google, when asked "one kilogram in
pounds" (without the quotation marks) responds with "one kilogram =
2.20462262 pounds".

Ticobol-ga seems to be arguing that since (most likely at least) if
you told someone that an object weighed 1 kilogram, that they would
know what you meant.  Therefore, kilogram is an accepted measure of
weight.  In this (limited) sense at least, he is correct.

Continuing Ticobol-ga's arguement on this is amount to: "How many
people need to use an incorrect definition for a word before it should
be consider a correct definition?"  This is more of a philosophical
question and doesn't need to be argued here.

In conclusion:
The scientifically correct answer is: Weight is a measure of force
(newtons) not mass (kilograms).
But in some ways, you are both correct.

-------------

Now for the 'work to tip over the chariot':

As already commented on, the problem itself is not very consistant.

For instance:
"It is going around a curve ...  Let's also ignore friction."
This is possible if we are going around a sloped surface (like some
racetracks have).

But there's also this comment, "...so there is a centripetal force
acting on it, parallel to the ground."  This is not true for a sloped surface.

Basicaly, without further comment from the question poster, we need to
make an assumption as to what he/she meant.  Everyone seems to have
agreed upon the assumption that the chariot is moving on a flat
horizontal surface with enough friction that it is not sliding
radially outward during the turn.

There is however, another slight assumption to be made, and I believe
this is where you two are disagreeing.

What work is the poster asking for?

For instance: Given the above situation, if the chariot is moving fast
enough, it will tip over.  No one had to run out and push it over. 
What work therfore are we referring to?

If the chariot is a uniform block (another assumption), then hegie-ga
/ racecar-ga's answer is to: How much work against gravity is needed
to bring the chariot to the tipping point?

I am not fully understanding Tricobol's argument, but I think it is
along the line: Yes, but what about the centripetal force or friction?
 Maybe he is trying to calculate how much work against gravity, the
centripetal force or friction is needed.


Let's start simple: the chariot is not moving.
There are only two forces: gravity and friction
Friction is acting on the side/wheel that we're tipping onto.  If I
choose my axis of rotation as this edge contact on the ground, then
friction provides no torque. (Because friction acts at the contact
point, not at the center of mass.) Only gravity will provide torque
about this axis.  If I calculate the total work to reach the tipping
point (integrate torque*angular displacement) I will get
hegie-ga/racecar-ga's answer.  (NOTE: This will work regardless of
where I choose the axis of rotation for defining the torque, but this
choice makes it so I don't have to solve for the friction forces.  ie.
it is the simplest choice for this problem.)

Now, let's have the chariot move.
There are again, __only two forces__ : gravity and friction.
This is important, very important.
"Centripetal force" is not some magical force.  It is just the name of
the force causing the "acceleration towards the center" in this case
it is friction causing the chariot to accelerate towards the center of
the 'track' it is turning around.  As mentioned earlier, friction acts
at the contact point, not at the center of mass.  Therefore the
calculation of the work is the same as above and gives the same
answer. Therefore, even if we calculated the friction force (which
would be different because it now includes the necessary 'centripetal
force'), the work to reach the tipping point is the same.

Therefore,hegie-ga/racecar-ga's answer is correct when the chariot is
moving as well.

Since this can occur without someone "running out and pushing over the
chariot", where does this work come from?  Since there are only two
forces involved: gravity and friction.  The work against gravity must
be done by friction.

So, if Ticbol-ga is trying to calculate the net work done against all
forces, the answer is zero (when worded as "net work done against all
forces" this should probably be fairly intuitive).

I hope that helped.

Ooops... I see Hedgie-ga already mentioned the "weight in space"
explanation.  He must have posted while I was typing.

Also, I forgot to mention in my above calculations that there is a
normal force. (The force preventing the chariot from falling through
the surface.)  Since this acts on the contact point, this does not
change the calculations, so I ignored it.


The more I think about it, the poster probably eventually wanted: 
Given a particular situation of the chariot, how much external work is
needed to tip it to the equilibrium point?

Which isn't really answered here (or even helped along with our
interpretation of the question).  Considering that the question itself
wasn't very consistent, they probably needed help working through the
concepts as well.

...maybe that's why the poster hasn't rated the answer yet.

snapogramer-ga, 

Your comments is full of holes. Too bad I have to go to work in a
while. I will comment on your comment in about 12 to 14 hours fromk
now.

(Original question-poster here:)  I didn't this posting would cause so
much arguing!  I neither have a physics degree, nor do I teach
mechanics, so I've been quite timid about responding.  It seems,
however, that some clarification is needed.

Yes, there is a frictional force to allow the chariot to round the
curve, to accelerate toward the centre of the circle he's describing;
and for simplicity's sake (for now, at least) I'm ignoring the effect
of skidding. Ultimately I just want to figure out how fast (velocity),
how tight a curve (radius), how long (time) and/or how far (arc
length) a charioteer could go without flipping over, i.e., without
rotating beyond the point of equilibrium.  If he rose up and then
righted himself, so be it!  In order to do that, I thought I would
first need to know how much work (in joules, please) was required to
bring the system to a point of equilibrium.

The way I saw it, W=Fs, F=ma, s=r x theta (radians), but in a rotational mode,
so Work = mass x arc length x angular acceleration
s = r x theta
  = 1.14 x .66 rad    (37.9 deg. convert to .66 rad)
  = .754

W = 70 x .754 x a (angular accel in radians per second squared)

Given constant velocity and radius (in rounding a corner) there will
be angular acceleration, provided the centrifugal force is enough to
overcome the righting torque of 480 newtons.  (How many blunders have
I made so far?  I better stop now.)

I haven't rated the question on Google's site because you've been
wonderfully generous at defending two or three different answers. 
Thank you!  My apologies if the question wasn't clear.

tharrop

In the metric system the unit of volume is liter, 10x10x10 cm, or
fractions and multiples thereof.  4 centiliter is a good shot of
whisky.    One liter of water weighs (or, if you will, masses) 1
kilogram (forgetting slight variances in gravity).  1000 liters is a
cubic meter (and a m^3 weighs one metric ton, which is why a metric
ton is heavier than our 2000 lb ton.
A newton  "is the standard meter-kilogram-second unit of force, equal
to the force that produces an acceleration of one meter per second per
second on a mass of one kilogram."  (the double "per second per
second" is correctly copied)
I don't know what that means ( I think the newton was introduced after
I left school.)
Here in Europe, people do not think of kilogram as a expression for
mass.  Sure they  have a visual impression of a liter milk and a 0.75
ltr wine bottle, and a one kilo loaf of bread, which is obviously of
much greater volume.
(Not many people understand that a liter of gold weighs as much as a
heavy suitcase, however, something to recall the next time you see a
film where gold bars or wooden boxes full of gold coins are being
tossed around.)

Anyway, kilogram is a weight (or mass), defined by that of a liter of water. 
A newton is a force defined in terms of kilogram, distance (meter) and
time (second).  Weight is not a measure of force, as someone stated,
regardless of the significance of "1 kg = 9.8 newton" (which patently
cannot be true from the definition of newton).

Centripetal force is that force that keeps a body moving on a curved
path, countering the body's inertia to move straight ahead (no
acceleration).
Centrifugal force is not a true force, just an expression for the
observed impression that the body wants to flee from the point it is
rotating around.  Actually, it just wants to go straight ahead  - at
any moment on a tangent to the circle that the centripetal force is
holding it to.

HI Todd,
Thanks for checking back in - while I was trying to define terms.
I feel vindicated:  you are interested in the chariot's speed and the
radius of the track, that is: on a track of given radius, what effort
is needed to accelerate Ben Hur until his chariot tips over.  Right?

Don't you agree that we need the radius of the track as a constant?

Hello snesprogrammer-ga

 and welcome to our family quarrel about mass and weight.
 When people with physics degree disagree about the meaning 
 of these terms, how can we expect the general public to cope?

You say:
"The problem is, since everyday life is here on earth, people
colloquially refer to kilograms as a measure of weight..."

By that you flat contradict my observation:
 When people say 
                    "I want 2 kg of apples" 
they want certain MASS of apples. They do not care what them apples weigh.
 They want to eat them and it is the mass which counts, which provides 
the calories (eh sorry- I mean Joules).

Ergo, people correctly use kg as unit of mass, when buing groceries.
I am with the people! They (the people) improperly abreviate kg as kilos,
not realizing they are asking for two thousand of apples. Since grocer
does not know that either, they use the same terminology, same system, and all
is well -- it is all just a convention, right?
So, issue is not with kg being unit of mass. Everyone, physicists,
people and even Tibolc agrees on that. The issue is this:
 
Do we mass them apples or do we weigh them? 

In either case, we put them on the balance or scales (a instrument
which in some languages is called 'weight') and using the force of
gravity (dubbed weight) we determine the mass. We express that mass in
kg. If we use weights (pieces of metal labelled by their mass) than
this procedure works as on Moon as well as it does on the Earth. If we
use spring scale, we need to recalibrate the scale. That's all -
except for the fact that apples are very expensive on Moon. It is even
worse then Hawaii or North pole, I am told. Everything needs to be
imported!

To bring in pounds (or even pounds of force, or even the obsolete kgm and kgf)
does not help to bring the debate to the end. Let's all agree and bring it to end!

 myoarin-ga

 Yes. You were right - asker wanted to know at what speed will the chariot tip
 over. So, he cannot eliminate friction. I had to introduce a boulder
to prevent the chariot from skidding.
That's out now, when cariot is moving. But you know, we are not able
to read minds, yet (I suppose you can?).
If  Tharop-ga  asks for clarification (in the RFC box) I will try to get that
expression. He needs not be shy since as a customer, he is in the driver seat. 

So the clarified question sems to be:

If chariot (modelled by a uniform box - on wheels) travels (without skiding)
on the horizontal track,
with speed v and 
track has constant curvature k  - as defined in
http://answers.google.com/answers/threadview?id=509922
then
 What will be the angle TAU, at which the chariot will be tipped?
 Which angle (and threfore speed) are critical (box just going to roll over)?

The tipping angle  TAU, the angle of the base with respect to the
horizontal ground, becomes critical whan the diagonal of the box
becomes vertical.
 That is easy to calculate from tg(side1/side2)= angle of diagonal to the base.

Ergo, what then is needed, is  this dependence of TAU(v,k) which
expresses the balance of the torgues. Any objections to this
formulation?

Hedgie

Edgeeh-ga,

You buy the mass of the apples? Or you just buy the apples?

"How much will be the mass of these apples, please?"
Is that not weird? Here on earth or on the moon?

What you do is buy the apples at the cost of their weight.

Or, If I follow your reasoning, in metric system you buy the mass of
the apples, while in English system you buy the weight of the apples.
You must have a very good brain, or your brain must be very good, to
switch to conditions per location of the stores.

Say, n apples that weigh 22 lbs in English, and equivalently, 22/2.2 =
10 kgs in metric.
If the a store uses English, you see the n apples in their weight of
22 pounds. If the store uses metric, you see the same n apples in
their mass of 22 kgs?

Weird, isn't it?

Oh, so in some places in Europe they are using/introducing that convention now?
Weird.

How do I like the new convention?
Weird.
Or, I will not even think of going in those European places.

The poor Europeans in those places then would be confused when they
have to buy apples is places/countries ehere apples are sold in
pounds, not in "mass".
Well, maybe, for those poor Europeans, a special/dedicated calculator
for apples should be a must when they stray away from their places and
they have to buy apples from the strange places' stores. Or, if the
calculators are not available, those poor Europeans might just cancell
going to such strange places.

That is just for apples.

What about in angling/fishing?
"Look, Pop, I just caught a snapper that weighs about 5 pounds,...er,
wait, do they use metric system in this country? ....I just caught a
snapper whose mass is about 2 kgm."

"When people with physics degree disagree about the meaning 
of these terms, how can we expect the general public to cope?"

Wow, this is weird...
We're not disagreeing on the scientifically correct answer. 
(As we obviously argee on that, as everyone should since it is 
not disputable.)  But we are disagreeing on whether or not 
people "colloquially refer to kilograms as a measure of 
weight".

Haha... this cracks me up.
I figured since several people here seem to have that trouble 
(and, as mentioned, google itself does) that indeed many people do 
colloquially refer to kilograms as a measure of weight.  But, 
I have not travelled very far (let alone throughout Europe), 
so my sampling of people is quite small...  So I will take 
your word for it, and quickly/easily concede.

-------------------------------------

tharrop-ga wrote:
"The way I saw it, W=Fs, F=ma, s=r x theta (radians), but in a 
rotational mode, so Work = mass x arc length x angular 
acceleration"

Sorry, that doesn't quite work.
Hopefully I can help a bit...

That equation will only be true if the forces in question are
tangential to the circle of radius r that you refer to.
This is because the first equation should be W = F dot product s 
(where F and s, are the vectors for the force and displacement
respecively).  If the angle between F and s is changing, then 
even this equation is only correct for each infintesimal displacement, 
ie  dW = F dot product ds ... and you'll need to integrate 
over the path to get the total work.

For example you wrote:
"s = r x theta
  = 1.14 x .66 rad    (37.9 deg. convert to .66 rad)
  = .754
W = 70 x .754 x a (angular accel in radians per second 
squared)"

The force of gravity is not tangential to the "circle" 
(which it appears you defined as the motion of the center
of mass during tipping), at any point in the rotation.  
Also, since the direction of the 'displacement' is changing 
during the motion of the center of mass, you'd have to
integrate dW = F dot product ds  to get the answer here 
(instead of just  F dot product s  from intial to final 
position).

-------------------------------------
Onto the question!

"Ultimately I just want to figure out how fast (velocity),
how tight a curve (radius), how long (time) and/or how far 
(arc length) a charioteer could go without flipping over, 
i.e., without rotating beyond the point of equilibrium."

For just going around a circular track at a constant speed:
let v = velocity around track
r = radius of curve/track
m = mass of chariot

h = initial height of center of mass (from the ground/track)
w = initial horizontal distance of center of mass from the tipping wheel

Going into the frame of the chariot's contact point with the ground
(which is an accelerating frame, so a fictitious force is introduced).
 In this frame the contact point is stationary, and the fictitous
force is "m*v^2/r" radially outward (from the axis of the 'circular
track').

In this frame, the torque about the contact point is:
Torque due to gravity = -mgw
Torque due to fictitious force = +(mv^2/r)*h
Torque due to normal force can be ignored here if the box is tipping
onto the contact point.

Past this point the magnitude of the torque due to gravity will become
smaller, and that of the fictitious force will become greater.  So if
it starts tipping it will continue tipping (on a circular track at a
constant speed at least).

The chariot will tip when mv^2h/r > mgw

Notice that the answer is independant of mass:
v^2/r > gw/h (which is some constant for your chariot)


as for the time:
let b = sqrt(h^2 + w^2)  (exactly what you were calling r earlier)
I = moment of inertia of chariot (depends on mass distribution in chariot)
theta = angle of center of mass from the ground


Torque due to gravity = -m g b cos(theta)
Torque due to fictitious force = +(mv^2/r)* b sin(theta)
Torque due to normal force can be ignored once weight is on the wheel edge

total torque: m((v^2/r) b sin(theta) - g b cos(theta)) = I (d^2 theta / dt^2)

You can integrate this twice to get the time it takes to tip from one
angle to another.  Unfortunately, this looks like it will be a nasty
integral, so I'll stop here.

Note however, that since the moment of inertial is always proportial
to m, that the "time to tip from one angle to another" is also
independant of the mass.

----------------------------------------

I must note:
Now that I have worked the problem this way, I believe I am incorrect
in my reasoning about "the moving chariot" case in my first comment. 
The frame of the contact point is an accelerating frame, so several
comments I made there are incorrect. (The "stationary chariot" case is
of course correct.)

[Ticbol, if this is what you're going to post about, please don't
bother. We shouldn't be filling this person's question responses with
arguements.]

I believe what I posted now is correct, but considering I've already
goofed earlier ... I probably should shut up in case I'm forgetting
other important details.

Besides, tharrop-ga is in good hands with racecar-ga and Hedgie-ga.

Sorry,...10kgs, not 22 kgs for the 22 lbs n apples.

Hello, Snuffygoramar-ga, I am back.

Before anything else, have you read my comment above on AjeiH-ga's
latest comment? About kgm not a unit of weight?

If not yet, go read it first, or later.

-------
" 'For instance:
"It is going around a curve ...  Let's also ignore friction."
This is possible if we are going around a sloped surface (like some
racetracks have).'"

No. 
Even on sloped racetracks, friction cannot be ignored. 
Without friction there, the cars/vehicles will fly off at the
slightest centrifugal force fighting the slightest centripetal force.

If no friction, how will the tires move? What will also hold the tires
from sliding radially outward?

--------
"But there's also this comment, "...so there is a centripetal force
acting on it, parallel to the ground."  This is not true for a sloped surface."

Eh?
You studied Physics, didn't you?

Do you mean there is no centripetal force?
Do you mean the centripetal force is not parallel to the ground?

You are wrong in both.

---------
"I am not fully understanding Tricobol's argument..."

Neat, Snoopyramegr-ga.

-------------
"Let's start simple: the chariot is not moving.
There are only two forces: gravity and friction"

Duh.
When there is no motion, or no impending motion, the friction is not yet activated.
If the edge is held in place by friction during the lifting, then more
than the gravity and friction are acting on the box. Whatever is
causing the lifting is acting on the box too.

"Only gravity will provide torque about this axis."

Nope.

"If I calculate the total work to reach the tipping point (integrate
torque*angular displacement) I will get hegie-ga/racecar-ga's answer."

No.
hegie-ga/racecar-ga's answer considered only the vertical component of
the total work.
The centroid travels horizontally also, aside from vertically, and
there is a horizontal force with varying strength equal to the varying
horizontal component of the friction at the edge on the ground. This
horizontal force pushing the centroid horizontally do work also.

Pushing, you may ask.
Yes, pushing. The centroid is at fixed distance from the egde. If the
tip of this fixed distance (where the centroid is located) is not
pushed towards the pivot's vertical plane, then this same tip will not
rise.
There is vertical work for the vertical travel of the centroid, and
there is horizontal work for the horizontal travel of the centroid.
The vertical force
due to the weight of the box and the horizontal force component of the
friction are the respective burdens.

--------------
"Now, let's have the chariot move.
There are again, __only two forces__ : gravity and friction.
This is important, very important.
"Centripetal force" is not some magical force.  It is just the name of
the force causing the "acceleration towards the center" in this case
it is friction causing the chariot to accelerate towards the center of
the 'track' it is turning around."

Zeez, Snipogarmo-ga, you really studied Physics?

"Therefore, even if we calculated the friction force ... the work to
reach the tipping point is the same. Therefore,hegie-ga/racecar-ga's
answer is correct when the chariot is moving as well.

You want me to cry, my friend?

----------
.........

"So, if Ticbol-ga is trying to calculate the net work done against all
forces, the answer is zero (when worded as "net work done against all
forces" this should probably be fairly intuitive).

Bo-hu-hu-hu.

--------------
"I hope that helped."

Waaaaaahhhh!

If I may say so, someone is not worrying about the problem in the question.
On a sloped track, and one curving up to the vertical, the chariot is
never going to tilt over if Ben Hur steers correctly, so that has
nothing to do with the question.
About friction, I thought there was a basic agreement that the problem
would be idealized: nothing skids and there is no effort calculated to
overcome friction, air resistance, whatever.

As snesprogrammer pointed out, perhaps the more relevant question is
how fast can you go around a turn without tipping over, rather than
how much work does it take to tip you.

R = radius of turn
g = accerleration of gravity
H = height of chariot
W = width of chariot
M = mass of chariot (used later)

The fastest you can go around the turn without tipping is v =
sqrt(RgW/H) (assuming infinite friction where the wheels contact the
ground, and zero rolling friction of course).   This answer was
already given by snesprogrammer and hedgie.

But here is something really cool: for small radius turns, you can
actually enter the turn faster than that, and make it through without
tipping over.  That is because as you enter the turn, the inside
wheels lift up as the chariot starts to tip.  As they lift, the center
of mass rises also, and thus some of the kinetic energy of the chariot
is converted into gravitational potential energy, slowing the chariot
down.  Of course, as the wheels lift, the torque needed to further tip
the chariot decreases, but if the reduction in speed is enough, there
will not be enough radial acceleration to tip the chariot the rest of
the way, even though it's easier to tip a chariot that has already
started tipping.

Fiddling around with the equations for potential and kinetic energy as
a function of tipping angle, here is what popped out:

If    R > W*H^2/(W^2 + H^2),

then the fastest you can enter the turn is v = sqrt(RgW/H).  That is,
if the radius of the turn is large enough relative to the size of the
chariot, then the fastest you can go is the speed for which the inside
wheels just barely stay on the ground.  If you go any faster, you will
tip.  As soon as the wheels begin to lift, you're done, splat. 
However:

if   R < W*H^2/(W^2 + H^2),

then you can enter the turn faster than v = sqrt(RgW/H).  Exactly how
much faster is tricky to solve for analytically.

Here's the process though.  If you assume that a chariot is rounding a
corner balanced on two wheels, then the total energy of the chariot is
made up of two parts--the kinetic energy (due to the motion), and the
gravitational potential energy (due to the fact that tipping the
chariot raises its center of mass).  In terms of the tipping angle X,
these energies are:

KE = MgR/2*([(W/H) - tan(X)] / [1 + (W/H)*tan(X)])

PE = Mg/2*(W*sin(X) + H*cos(X) - H)

Differentiating the total energy (KE + PE) with respect to X, and
setting the result equal to zero gives you an equation for the angle X
which corresponds to the maximum total energy.  This equation is:


***************************************************************
(W*cos(X) + H*sin(X)) * (H*cos(X) + W*sin(X))^2 = R*(H^2 + W^2)
***************************************************************


If R < W*H^2/(W^2 + H^2), this equation can be solved for X, which
then allows you to calculate the maximum speed v at which you can
enter the turn.  The equation is easy to solve numerically, but I
don't know how to do it analytically.

That is great analysis and neat expression,  racecar

      Now,   consider  right-angle triangle  H W 
               and diagonal D= SQRT ( W*W + H*H) 
              with Chariot-angle C : cos(C)= H/D and  sin(C)=W/D 
              
then you can divide whole eq. by D^3 to get

(sin(C)*cos(X) + cos(C)*sin(X)) * (cos(C)*cos(X) + sin(C)*sin(X)) ^2 = R/D

         or    cos( C + X) sin (C -X) ^2 = R/D 
           
              
 Does this has any geometrical meaning? 

What it is good for? Perhaps a nice textbook problem ??

Hedgie

Thank you, everyone, for diving in and offering pearls of thought. 
The debate has been very interesting, not to say entertaining. 
There's a wealth of information here to sift through.  I feel a little
cheap for asking the question since my motivation was so trivial, but
I hope the discussions were useful to all of you.  Good luck with your
endeavours!

Todd H.