3 things: Even after reading many reviews, I still can't fathom these...

[86BRZ]

Need to understand the 86 to love (and forgive) the 86.

[wu_dot_com]

Originally Posted by strat61caster

Wait, are you saying that lowering the car, (resulting in a chassis that sits lower to the ground) does not change the height of the Center Of Mass?

No arguments about how negligible the effects are for 99% of drivers, your post just doesn't make sense to me.

No, I'm saying lowering the car does not change the center of gravity (COG). Which depends on the context it's used, is not the same as COM.

[oldpueblo]

Driving is basically like this:

And here of course is a musclecar :

Now choose!

[wu_dot_com]

Originally Posted by ZDan

Oh yes it does! Lowering a car 1" will absolutely lower its center of gravity by almost 1".

These are the *same thing*.

Will gain you zero points in an engineering/physics/science test!

Center of gravity and center of mass both refer to a single point in space. The *same* point in 3d space. Neither actually describes the mass distribution, but only the place where for simple analyses the total mass could be said to reside.

No, center of gravity and center of mass only refer to a location. Neither is a "force" or a "mass". They are an address of a location represented in x,y,z or other 3d coordinates.

Of course lowering the car doesn't affect its actual MASS, or its WEIGHT in a gravitational field!

Lowering the car *does* move the c.o.g/c.o.m. location downward by very nearly the amount of lowering. This is very beneficial as it allows the inside tires to contribute more and will generally increase cornering capability for reasons having to do with the nonlinear relationship between tire lateral grip and vertical load.

Please do not discredit Wikipedia until you have better appreciation of the subject at hand.

Second, I am actually a engineer by trade with state certification in the subject of mechanical engineering.

Third COG and COM can only be the same point if the assemptions is to which the system of particles are infact a ridgit body. Depends on which level you are analyzing, COG and COM can be very different.

Also COG by definition where the resulting torque due to gravity in that given location equal zero. The COG equation is written as f(r)=-q(r)dVgk. -is the general noteation of gravity force direction. V is volume while q is density at each point of the total volume. g is gravity while k is the directional unit vector of the vertical(y) plane. Using sum of the moment equal zero or sum(T)= int(V) ((r-R)xf(r))=0 where R is the respective location of center of gravity of each x,y,z plane. In this case because your unit vector in x, and z plane is zero, your sum of moment would also be zero in y and z plane. As a result, COG is only useful in determining a reference location on the horizontal plane where the partical body system will not rotate due to gravity.

Therefore changing the ride hight of your car does not change the COG because it does not change the resulting torque on system body due to gravity.

No wonder our engineering and science student becomes less competitive on the globe market. It seems like they don't even bother to teach the distinction between the two anymore

[ZDan]

Center of mass, center of gravity are exactly as I described. Lower the car, lower the c.g., center of mass. Period.

[wu_dot_com]

Originally Posted by SubieNate

Sort of kind of maybe. Not.

"Center of gravity" on a car can be used interchangeably with "Center of Mass" as gravity acts on the car, effectively, at the center of mass.

That said, your statement that it doesn't matter because gravity acts in the vertical plane is false. While turning or making any kind of acceleration, the force acting on the car is a combination of gravity and and forces fore and aft and left and right as the car accelerates forward or backwards (braking) and to one side of the car or the other (angular acceleration) as the car turns.

Inertia wants to keep the car moving exactly how it is. It takes force to deviate from this. The imaginary inertial force effectively acts on the center of mass of the car and is reacted by the tires.

By lowering the car (properly) you reduce the roll moment induced by the force it takes to turn the car. This moment is reacted by your tires as an apparent decrease in load on the inside tires and apparent increase in load on the outside tires. By reducing this change in apparent force, you better utilize all of the contact area of all four tires, leaving more usable grip to actually turn the car.

Think about a truck vs our cars. Do you think that there is any way that a truck could ever corner as easily with the same tires our car has?

Now. The amount of change you're going to be able to affect may or may not be noticeable, or it may only be noticeable on the track. But there most definitely is a change.

Cheers
Nathan

I agree with your analysis on forces acting on the partical system center of mass.

Let me ask you this, when you set up your dynamic calculations, do you consider your acceleration as a directional vector sum or do you have a seperate equation which breaks out the effects of gravity?

Though both approach would technically be correct, but breaking out the resulting reaction due to gravity would be redundant.

If you use the directional vector sum, which as you describe in your post, then you are talking about COM.

During the vechicle dynamic analysis for external forces acting on the car, assuming the car is a ridgit body where the weight of the body is concentrated at COM is convenient. However when analyzing the internal particle system's reaction to each others, using COM would be more appropriate. this is because you can no longer assume the car behaves like a ridget solid body.

[u/Josh]

The only time center of gravity and center of mass are in different places is if the gravitational field cannot be considered constant over the whole body.


And since we are quoting wikipedia

Center of gravity is the point in a body around which the resultant torque due to gravity forces vanish. Near the surface of the earth, where the gravity acts downward as a parallel force field, the center of gravity and the center of mass are the same.

[wu_dot_com]

Originally Posted by ZDan

Center of mass, center of gravity are exactly as I described. Lower the car, lower the c.g., center of mass. Period.

ENGINEERED:happy0180:

[SubieNate]

If you think in three dimensions, you will have a "center of mass" in every axis. X (front to back) Y (side to side) and Z (up and down). For a rigid body analysis, that is the point you use when calculating the moments and forces applied to the tires by the mass of the car as it accelerates in any form.

Particle systems? The system is rigid enough and the amount of mass that is free to move is so small that doing an analysis like that is just foolish. The potential for increased accuracy would easily be taken up by the variation in test results caused by changes in driver input, random variation, noise in the measurement system, etc. This isn't particle physics. Things on a macro scale (E.G., a car) are way too complicated and there are far too many "particles" to even consider using that type of analysis for any kind of practical problem solving.

The manner in which you analyze the system, whether you do a sum or some sort of superposition when it comes to gravity should have no net effect on the final result. The physical system is the physical system. Any analysis that does not net the same final answer (disregarding differences in the assumptions and accuracy of the type of calculation. Assuming mathematical method is the only change) is wrong.

Cheers
Nathan

[Brzetto]

The BRZ/86/FRS is not slow, the other cars just have a lot more power to compensate for their relatively humongous weight.

[wu_dot_com]

Originally Posted by u/Josh

The only time center of gravity and center of mass are in different places is if the gravitational field cannot be considered constant over the whole body.


And since we are quoting wikipedia

Not disputing that fact. But here is the question, how can you accurately setup your sum of moments (torque) equal zero for all 3 planes to solve for R when your unt vector in X and Z direction is zero?

[fistpoint]

Originally Posted by lemonspeakers

1) When people say this car is slow. How slow am I talking? Can you put it into perspective of a 1998 Honda Civic SI or a 1999 Acura Intega GSR?

There was no '98 Civic Si...but to answer your question: the FR-S/BRZ is 98%+ identical in performance to not only the latest generation of Civic Si, but the previous one as well(there were no number gains even with the larger engine '12 model).

The funny part is that when you goto the Civic forums they all yell about how slow the twins are, even after you present them with well thought out and sourced links showing them the same results from the same car magazines for all 4 versions of the cars in question(FR-S/BRZ, Si 2 and 4 door). 1/10th faster to 60mph for one, 1/10th faster for another, trading blows back and forth with no clear winner.

[u/Josh]

Originally Posted by wu_dot_com

Not disputing that fact. But here is the question, how can you accurately setup your sum of moments (torque) equal zero for all 3 planes to solve for R when your unt vector in X and Z direction is zero?

I see what you are getting at and I'll try to explain, but the whole point is moot because if you want to be absolutely accurate it is the lowering of the center of mass of the car that improves performance when the car is lowered.

I think the point you are trying to make is that if you take the (x,y) points of the center of mass, you can take any arbitrary z value and the moment about the center of mass due to gravity force about that point will be zero if you assume the gravity force is only in the z direction. However, we don't define center of gravity by only (x,y) because while that is useful when the car is on flat ground and gravity only acts in the z, imagine the situation where the car is parked on a hill. Assume the z plane still goes up through the roof of the car, gravity force acts partially in the z direction and partially in either x, y, or both. Now there is only one (x,y,z) position where the moment about the center of mass due to gravity force is zero.

[wu_dot_com]

Originally Posted by u/Josh

I see what you are getting at and I'll try to explain, but the whole point is moot because if you want to be absolutely accurate it is the lowering of the center of mass of the car that improves performance when the car is lowered.

I think the point you are trying to make is that if you take the (x,y) points of the center of mass, you can take any arbitrary z value and the moment about the center of mass due to gravity force about that point will be zero if you assume the gravity force is only in the z direction. However, we don't define center of gravity by only (x,y) because while that is useful when the car is on flat ground and gravity only acts in the z, imagine the situation where the car is parked on a hill. Assume the z plane still goes up through the roof of the car, gravity force acts partially in the z direction and partially in either x, y, or both. Now there is only one (x,y,z) position where the moment about the center of mass due to gravity force is zero.

Very good point.

I think I've turn 90% of people away from this topic with this geek talk.