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## Force causing rotation!

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### #1uncutno  Members

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Posted 21 June 2001 - 03:05 AM

(Don''t mind bad english!) Ok, here is my question! If a force is added to an objekt. How mouch will cause the object to rotation, and how mouch to movement(accseleration)? Because if u use the model where a force-vector paralell to the vector betwen the hit-point, and the center-mass-point would only cause movement, and a force-vector diagonal to the vector betwen the hit-point, and the center-mass-point, would only cause rotation, You would get a problem, when to exact simular forces was added diagonal to the hit-points/center-mass-points, like this: h---->f | | 0 | | h---->f h=hitpoint, f=force O=center-mass-point THis would cause the object to try to rotate two ways, witch would sum up to a 0 rotational accseleration: No movement, no rotation! ???? Is there a simple formula? like: v=at or F=ma to this problem?? -Anders-Oredsson-Norway-

### #2Bandures  Members

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Posted 21 June 2001 - 04:43 AM

M = FR
Where R = distance from center mass
M - moment ( i dont''t know how it in english )
If sum of M = 0 , then no rotation
M=I*w*w , where w - Angular speed , and I - moment of inertia ( const ) - it''s describe mass distribution in object.

### #3grhodes_at_work  Members

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Posted 21 June 2001 - 05:37 AM

quote:
Original post by uncutno
Because if u use the model where a force-vector paralell to the vector betwen the hit-point, and the center-mass-point would only cause movement, and a force-vector diagonal to the vector betwen the hit-point, and the center-mass-point, would only cause rotation

Your first observation is correct: if the force vector is parallel to the vector between the hit point and center of mass, the force will not cause rotational acceleration. The reason is that the force causes no torque (moment-of-force) about the center of mass, and it is torque that causes the rotational acceleration.

Your second observation is incorrect: an unbalanced force will *always* cause movement/translation, whether the force is parallel or diagonal (did you mean to say normal/perpendicular).

quote:
Original post by uncutno
, You would get a problem, when to exact simular forces was added diagonal to the hit-points/center-mass-points,
like this:

[diagram removed by grhodes_at_work]

THis would cause the object to try to rotate two ways, witch would sum up to a 0 rotational accseleration: No movement, no rotation!

Your observation here is partially correct: it is true that the two forces try to cause rotational acceleration in opposite directions, leading to 0 net rotational acceleration. That is because one force causes a positive torque and one causes an equal but opposite negative torque. The sum of all torques equals 0 net torque.

It is not true that this system will experience no movement/translation. Your two forces, f, do not cancel each other out. Both applied forces, f, are acting in the same direction----to the right. So, in reality the net applied force is equal to 2 * f to the right. The object will undergo translational acceleration to the right.

quote:
Original post by uncutno
Is there a simple formula? like:

v=at or F=ma to this problem??

Yes, certainly, and F=ma is one of them. For a simple point mass representation of the object (which is what you''re describing), use two basic equations. Its best to stick to 2D physics until you''re comfortable with the way forces and torques work, and I present these equations for 2D. They do extend to 3D, but things do get messy pretty fast---there are additional terms in the equations, etc. You should be comfortable with the 2D first.

Equation (1) is to determine movement/translational acceleration.

(1) F = m * a for translation, where F is the force vector, m is the mass, and a is the acceleration vector. There are actually two equations, one for the x direction and one for the y direction: F_x = m * a_x and F_y = m * a_y.

Sum the force vectors up, and solve for a_x and a_y at each time step. You can then integrate dv/dt = a to get the velocity at the new time step and then once you have velocity integrate ds/dt = v to get the new position. velocity v and position s are two-dimensional just like v.

Equation (2) is to determine rotational acceleration.

(2) T = J * ra, where T is the net torque, J is the moment of inertia, and ra is the rotational acceleration (radians/sec_squared). There is only one equation for 2D problems: the T and ra vectors are always perpendicular to the 2D plane.

Sum the torques up, solve for ra at each time step. Then, integrate drv/dt = ra to get rotational velocity, rv, and drs/dt = rv to get new rotational position, rs.

Chris Hecker''s resources on the web (especially the Game Developer Magazine articles from a few years ago) are a good beginner''s resource for how to do all this in detail.

Of course, if you do have collisions and things, it gets more complex even in 2D, but perhaps this will give you a start.

By the way, the equation "v=at" is actually part of a closed form solution for projectile motion. It is valid for objects undergoing gravitational acceleration with no other forces applied.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

### #4grhodes_at_work  Members

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Posted 21 June 2001 - 05:53 AM

quote:
Original post by Bandures
M=I*w*w , where w - Angular speed , and I - moment of inertia

This equation is wrong, by the way. In 2D, it would be:

M = I * dw/dt, where dw/dt is the time derivative of w, or the angular acceleration.

In 3D, it would be:

M = I * dw/dt + w cross (I * w), where here I is the 3x3 inertia tensor matrix and w is a vector.

I see where you may have made your mistake. If you naively simplify the 3D equation down to 2D, so that I and w are scalars, you would get:

M = I * dw/dt + w * ( I * w)

or,

M = I * dw/dt + I * w * w

which looks just like your equation with the (required) term I * dw/dt added.

But there's a problem with that: w cross (I * w) does not simplify to w * (I * w). If you notice that for 2D problems, w = (0, 0, wz) and I is a diagonal matrix (i.e., no inertial coupling), then expand out w cross (I * w), you will see that the term is 0 for 2D problems and you are left with my equation for 2D.

Oh, by the way, the word "moment" is perfectly fine in English! We also call it torque.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

Edited by - grhodes_at_work on June 21, 2001 12:54:44 PM

### #5furby100  Members

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Posted 21 June 2001 - 09:24 AM

v = at?
Surely you mean v = u + at......

### #6uncutno  Members

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Posted 21 June 2001 - 10:14 AM

Ok.. thanks! I will try to make this into my engine, and by the way, its 3D :-) I earlier tried to simulate this behavor by making a model of the car as 8 points in a cube with elastic springs betwen... Im making a remote-controled-wipeout-sortof game, and i need some better physics then [right_key] => angle+=k; :-) like you se in so many bad games...
I will surly post a message when the game is done. :-)
and pleas send e-mailbombs to the norwegian state, for not teatching this in scool! :-)

anything more? For example, if you slide a car into a corner, how do you calc the force that will prevent the car from "entering the corner, and make it rotate instead? I have seen some of these formulas before, but never a practical implantation...

p.s: my assumtion that u(start velocity) was zero, could cause alot of bad thing! It will never happen again! :-)

-Anders-Oredsson-Norway-

### #7grhodes_at_work  Members

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Posted 21 June 2001 - 10:36 AM

quote:
Original post by furby100
v = at?
Surely you mean v = u + at......

Yes, of course it is this, with u being the initial velocity for the constant acceleration case.

### #8 Anonymous Poster_Anonymous Poster_*   Guests

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Posted 21 June 2001 - 12:45 PM

Now THIS is a physics question!!

quote:
Original post by uncutno
For example, if you slide a car into a corner, how do you calc the force that will prevent the car from "entering the corner, and make it rotate instead?

The idea is exciting, and I think it''s readily achievable by simple playtesting. You want to be able to assign friction (or, more precisely, lateral acceleration) constants to each of the tires, have either total or distributed mass (if you want to get REALLY sexy, you could simulate the suspension as four separate masses, each near the tire, just the way a real car works, but hey), and on the motion side you want your vehice''s velocity vector, and the angles of the road the vehicle is on/approaching/exiting.

When I first envisioned the problem, my mind immediately leapt to the car chase at the end of "Bullitt" with Steve McQueen. Nash Bridges is a half-ass attempt at being Bullitt-like in driving style, but it''s only an attempt.

I don''t know the formulae, but if the vehicle is moving with too much energy for the lateral acceleration coefficient of the tires, the vehicle will spin OUT if it attempts too sharp a turn (probably around 40 or 50 degrees incidence to the velocity vetor, I would guess).

In order to spin IN to a turn, the drive tires (I say it this way because it could be RWD, FWD, or AWD) would have to be sufficently worn down, and the front tires (always the front) would have to bite in pretty hard. Unless you have some complex vehicle physics, tire physics, etc, this may be a tough nut to crack.
-----------------
-WarMage
...what, me be technical?

### #9flame_warrior  Members

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Posted 21 June 2001 - 01:03 PM

Hi,

Use gyroscopic effect to figure out how much the car should turn. Lesser calculation and more accurate results.

### #10WarMage  Members

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Posted 21 June 2001 - 01:11 PM

flame_warrior, I''m not sure I''m convinced. How would you calculate the point at which to break the tires loose, and the angular momentum associated with it? If it works, that sounds cool (I did post the somewhat longwinded one above), but could you give a little more background to your solution?
------------
-WarMage
...I used to give a rat''s ass, but I ran out of rats.

### #11Bandures  Members

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Posted 21 June 2001 - 09:22 PM

2grhodes_at_work:
quote:

This equation is wrong, by the way. In 2D, it would be:
M = I * dw/dt, where dw/dt is the time derivative of w, or the angular acceleration.

M = FR = mar
where a center orientaited acceleration ( i mean tan acceleration = 0 and w = const )
a = w*w*r
M = m*r*r*w*w , get I = mr*r
I think this will be all , what he need.
And where you see here "wrong" ?

If it come from plane to 3D space, certainly all you formulas will be true , but if it didn''t know moment at all , i don''t think it understandant your post

### #12grhodes_at_work  Members

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Posted 22 June 2001 - 06:31 AM

Since the original poster has answers to his questions, I''d like to resolve some final issues and then close this thread. The sub-topic regarding race cars spinning out (another message) should be continued in a new thread if there is interest.

quote:
Original post by Bandures
M = FR = mar
where a center orientaited acceleration ( i mean tan acceleration = 0 and w = const )

I see a number of possible problems with your analysis here, but before I reply I''d like to clarify my understanding of your variables. It will help me better understand what you''re trying to do.

Question 1) Do you mean for R = r = the distance away from the center of mass where the new force (F) is applied?

Question 2) When you say a is the "center orientaited acceleration," do you mean for a to be the centripetal acceleration of the point on the object surface where the force F is applied *OR* do you mean for a to be the translational acceleration of the center of mass of the object?

I understand that dw/dt = 0 (and tangential acceleration also = 0) when w = const.

quote:
Original post by Bandures
a = w*w*r

This equation implies a particular answer to question #2.

quote:
Original post by Bandures
M = m*r*r*w*w , get I = mr*r

Question 3) In your equation I = mr*r, do you mean for "r" to be the same as the "R = r" from question #1?

Please answer my questions, and then we''ll be able to settle our differences!

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

### #13flame_warrior  Members

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Posted 22 June 2001 - 07:02 AM

Original Post by WarMage.

flame_warrior, I''m not sure I''m convinced. How would you calculate the point at which to break the tires loose, and the angular momentum associated with it? If it works, that sounds cool (I did post the somewhat longwinded one above), but could you give a little more background to your solution?

Hi,
Its tough for me to explain here. It cannot be explained without the use of drawings(by me anyway). I can point you to some books where you can find this info though. Try to find some books on Theory of machines which has Gyroscopic Effects and Precessional Motions. Any Mechanical Engineering(studying in 5th or 6th sem) Student should have it.

I can give you the equations for overturning though if you want it that is.

### #14Bandures  Members

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Posted 22 June 2001 - 09:33 AM

1) yes , r - radius of inertion ( all mass on this radius )
2) "the centripetal acceleration of the point on the object surface where the force F is applied"
Ofcouse F = F applied only if we have fixed center.

### #15grhodes_at_work  Members

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Posted 22 June 2001 - 11:45 AM

quote:
Original post by Bandures
1) yes , r - radius of inertion ( all mass on this radius )
2) "the centripetal acceleration of the point on the object surface where the force F is applied"
Ofcouse F = F applied only if we have fixed center.

Okay, excellent. This clarifies your assumptions very well. There may still be some missing information, and a picture would immensely help but I can clarify a few things with what we have. I''m afraid I still see inconsistencies in your approach. First of all, the correct standard terminology for your "r" is "radius of gyration", not "radius of interia/inertion."

Lets go back to your previous post:

quote:
Original post by Bandures
M = FR = mar
where a center orientaited acceleration ( i mean tan acceleration = 0 and w = const )

Two things are inconsistent in this part of your post. First, if you allow M to be nonzero, then the tangential (and rotational) accelerations would also be *nonzero*, leading to w being *not* constant----in conflict with your stated assumption. This is because M *causes* rotational/tangential acceleration. Thus, given your assumption that tan acceleration = 0 and w = const, it would be impossible to have a nonzero M. Second, you have stated that a is the centripetal acceleration. By writing M = mar you are relating moment, which causes rotational acceleration, to centripetal acceleration. Centripetal acceleration is radial and is orthogonal to rotational acceleration and cannot produce a nonzero M. Thus, M = mar is incorrect. (Your force, F = ma, is centripetal force.)

Lets look at one further part of your previous post:

quote:
Original post by Bandures
a = w*w*r

Well, of course this is an equation for centripetal acceleration, and as described above it cannot be related here to the calculation of M, and so your expanded equation,

quote:

M = m*r*r*w*w

is also incorrect.

It appears to me you have attempted to change equations for forces and accelerations related to an object moving in a perfect circle at constant speed into equations that can represent an object that may undergo arbitrary forces and moments about its center of mass. uncutno''s original question did deal with forces applied both radially through the center of mass causing translational acceleration and tangentially causing rotational acceleration. Simple constant centripetal acceleration isn''t applicable to that problem.

Does this clear up some misunderstandings?

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

### #16 Anonymous Poster_Anonymous Poster_*   Guests

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Posted 24 June 2001 - 04:40 PM

i think thier overcomplicating the answers to the question
in the second post he wanted to how to calculate when the car will slide in a turn

so i would say because the front wheels begin the turn and causes rotation as far as the center mass of the car is concered the driver turns the wheel to compensate and increase friction on the front tires the back tires do not turn as a result a micro vibrational effect occurs that decreases the grip friction of the back tires and the back of the car begins to slide in part from momentum and in part from rotation i have no idea how to put it into math but their it is

### #17 Anonymous Poster_Anonymous Poster_*   Guests

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Posted 24 June 2001 - 05:12 PM

to add this is the crucial part of the calculation because if the center mass of the cars momentum overcomes the (friction grip) of the tires it will slide into the wall
if the friction overcomes the gravity of the cars mass or the momentum the car may roll or may violenty reverse rotation

u could would also have to consider the fact that the ground affects friction water on the road the type of road the speed of the car the condition of the tires the shocks windflow other factors affect each tire in a turn obviously the inside tires have less force on them as the outside ones

u also have to take into consideration the fact that the higher the cars center mass is the more unevenly distributed the (force and weight) on the tires will be a car very low to the ground car will have a very evenly distributed amount of friction to each tire ect...

the turning of the cars front wheels generates the
rotation relative to the back wheels of the car the amount of
friction of the front tires to the back tires determines increaseing or decreaseing rotation once the the turn has begun

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