You do not really get the energy back, BUT as the car accelerates down the track, the wheel slipage diminishes to the point where all of the engines energy is effectively used driving the car forward, only!Originally posted by meister
How do you get the energy back when the front end drops back onto the ground? You use engine power to lift it up, then whatever potential energy you gained while in the air doesn't all go completely towards accelerating the car.
How is this different from a drag car that keeps all four wheels on the ground at all times?Originally posted by Mr. Robin Parsons
You do not really get the energy back, BUT as the car accelerates down the track, the wheel slipage diminishes to the point where all of the engines energy is effectively used driving the car forward, only!
No longer enough torque on the frame/chassis to lift it, so it transfers to the wheels to drive forward
Where does it go if not toward driving the car? It must go somewhere, it can't just disappear.Originally posted by meister
How do you get the energy back when the front end drops back onto the ground? You use engine power to lift it up, then whatever potential energy you gained while in the air doesn't all go completely towards accelerating the car.
There is a slight dip in the acceleration curve, but the overall average acceleration remains the same.How is this different from a drag car that keeps all four wheels on the ground at all times?
COG never shifts. Its a center of mass of the vehicle. Its weight that shifts, like every time you walk, your weight shifts from leg to leg. Your COG remains at same spot of your body.Originally posted by Mr. Robin Parsons
Although the COG shifts towards the rear when the car lifts, it is the torque application (wheels/axles to frame/chassis) that is lifting the car. Hence the manner of drive, axle, shaft, belt makes little (But not absolutley "NO") difference in the way it will cause lift.
The energy isn't really "lost", per say, but is not being used to drive the car forward as much as it is being used to lift the front end, counteracting gravity.
There is almost no energy consumed on lifting the car as it almost doesn't lift. The energy that seems to go to lifting the car front against gravity in reality goes to pressing rear wheels against ground. Its a weight shift, purely due to inertia and COG being above ground. Front end of car almost becomes weightless as full weight of car goes to rear end.How do you get the energy back when the front end drops back onto the ground? - meister
Originally posted by russ_watters
Where does it go if not toward driving the car? It must go somewhere, it can't just disappear.
So when I had mentioned the COG "shifting" I had meant 'relative to the ground' cause it does.Originally posted by wimms
(SNIP) COG never shifts. Its a center of mass of the vehicle. Its weight that shifts, like every time you walk, your weight shifts from leg to leg. Your COG remains at same spot of your body. (SNoP)
The lifting of the front end of the vehicule is due to the torque exerted by the driving wheels upon the frame/chassis of the car, not inertia.Originally posted by wimms
(SNIP) Its a weight shift, purely due to inertia and COG being above ground. Front end of car almost becomes weightless as full weight of car goes to rear end. (SNoP)
The way you put it is confusing and probably wrong - weight is a biproduct of mass, so center of mass and center of gravity (weight) are the same thing. MRP says it right:Originally posted by wimms
COG never shifts. Its a center of mass of the vehicle. Its weight that shifts, like every time you walk, your weight shifts from leg to leg. Your COG remains at same spot of your body.
Yes. Its the torque that lifts the front wheels. However, this is wrong:The lifting of the front end of the vehicule is due to the torque exerted by the driving wheels upon the frame/chassis of the car, not inertia.
The COG is simply a static physical property of the object.So when I had mentioned the COG "shifting" I had meant 'relative to the ground' cause it does.
Not quite. If the COG were directly over the back wheel, the car would flip over backwards at the slightest applied torque. What is required is having the COG as far back as possible to maximize the weight on the back tires, while making the torque required to lift the front wheels as high as necessary to keep the car from flipping over backwards.A "funny car" with its bicycle front wheels and cog all the way back is set up at a standstill in very nearly this condition.
And by this example, (above) we can see the the center of gravity shifts relative to the ground, simply by lifting the far end of the long pole (that we hold by only one end) and noteing that the COG, relative to the ground, approaches the end of the stick that you are holding.Originally posted by russ_watters
(SNIP) Take a yardstick or a broomstick or something long and skinny and hold it horizontally in one hand at the center. The center is the COG. Notice you are not applying any torque to hold it level. Any small torque and you can spin it.
Now hold it towards one end, again keeping it level. The COG hasn't changed and the overall weight is the same, but now you need a pretty large torque to keep it level.
(SNoP)
While COG and COM are same, center of distribution of weight is not. Weight is meaningless without support, and it does depend on acceleration forces. If a-force were applied to a line parallel to ground and at height of COG, no front lift would occur.Originally posted by russ_watters
weight is a biproduct of mass, so center of mass and center of gravity (weight) are the same thing.
He seems to imply that its the angular momentum of rear wheels that causes front lift. Its not. Although angular momentum is conserved, its impact is way too small. Just compare angular momentum of wheel radius and their mass to that of chassis radius and its mass. Besides, angular momentum causes car to turn around its COG, not rear axle. This means that angular momentum forces rear axle to move towards ground again, being stopped by it.MRP says it right: Yes. Its the torque that lifts the front wheels.
By torque, you mean accelerative force, not angular acceleration of rear wheel, right?The COG doesn't change, but where the COG is affects the torque required to lift the front end.
This is a dynamic event and requires that you examine the COG relative to something, the ground! and relative to the ground it moves, hence the weight transfer.Originally posted by wimms
(SNIP) Robin, COG NEVER moves. Max you can think of is movement of suspension and taking this as change in goemetry of vehicle. (SNoP)
COG is defined as center of mass relative to geometric bounds of vehicle, not to the ground. And car moves forward, of course COG as point moves. But as long as mass doesn't move inside a car, COG does not move within it.Originally posted by Mr. Robin Parsons
This is a dynamic event and requires that you examine the COG relative to something, the ground! and relative to the ground it moves, hence the weight transfer.
rather off, huh. Then show me how.Forgive me, but the rest of what you write seems rather off.
Yes, but in this instance we must adjudicate the effects upon the car as it is relative to the ground, otherwise we will not even know that it has lifted up.Originally posted by wimms
(SNIP) COG is defined as center of mass relative to geometric bounds of vehicle, not to the ground. And car moves forward, of course COG as point moves. But as long as mass doesn't move inside a car, COG does not move within it. (SNoP)
This statement makes some kinda strange non sense to me. It does have some sense to it, but it is NOT applicable to what the "original question asked" is about, as clearly we are NOT dealing with something floating in space we are dealing with something that is on the ground, and is affected by gravity, and is acting because of TORQUE that is being exerted upon the parts, not the angular momentum of the entire chassis which is what you seem to be implying in this quote...Originally posted by wimms
Imagine this car in space. When you apply force to it, it either simply moves, or also starts rotating around its COG, depending on where the force is applied. In our case, front lift is not due to engine torque fighting gravity, but because car rotates around its COG. This rotation is stopped by rear wheels at ground, and this translates into weight shift, almost 100%. This is pure inertial behaviour. Weight of car never changes, nor is its COG lifted higher above ground. Thus, gravity is not beated. Its the only force we have to keep that car on the ground (besides aerodynamics).
The angular momentum of the wheels, and the chassis, are relitivised by the exertion of torque through the pinion and the ring (King and Crown, old school) which is what allows for the overcoming of the forces as to cause the front end to lift.Originally posted by wimms
He seems to imply that its the angular momentum of rear wheels that causes front lift. Its not. Although angular momentum is conserved, its impact is way too small. Just compare angular momentum of wheel radius and their mass to that of chassis radius and its mass. Besides, angular momentum causes car to turn around its COG, not rear axle. This means that angular momentum forces rear axle to move towards ground again, being stopped by it.
And why do you think that COG lifts? Are you saying that with sudden enough acceleration force, the car will jump off the ground with all 4 wheels??Originally posted by Mr. Robin Parsons
Yes, but in this instance we must adjudicate the effects upon the car as it is relative to the ground, otherwise we will not even know that it has lifted up.
First, this wasn't polite. Second, are you claiming that inertial effects that occur in space do NOT occur in gravity?? Perhaps you've forgot that we have to do with acceleration not constant torque applied to stationary object. Perhaps you've forgot that parts that take the torque are located all way back behind COG. Perhaps you forgot that with 3G acceleration inertial effects become increasingly important.Originally posted by Mr. Robin Parsons
This statement makes some kinda strange non sense to me. It does have some sense to it, but it is NOT applicable to what the "original question asked" is about, as clearly we are NOT dealing with something floating in space we are dealing with something that is on the ground, and is affected by gravity, and is acting because of TORQUE that is being exerted upon the parts, not the angular momentum of the entire chassis which is what you seem to be implying in this quote...
Seems you aren't relativizing enough. From your post it seems you imply that chassis lift occurs around rear axle. But you can't ignore inertia of car that is the only reaction force, concentrated to COG, which is located above rear axle and in front of it. Resultant acceleation force vector is parallel to the ground, any other idea would violate some conservation law. COG cannot depart from ground, as this would be liftoff of the whole mass of the car and reduction of traction. The only possibility that remains is force vector directed ahead between COG and ground, causing chassis rotation around its COG.The angular momentum of the wheels, and the chassis, are relitivised by the exertion of torque through the pinion and the ring (King and Crown, old school) which is what allows for the overcoming of the forces as to cause the front end to lift.
Although for our case its effect is small, your comment is wrong unless I've again misunderstood something. Car together with its wheels forms system. Angular momentum is conserved for the whole system.The 'angular momentum' imparted to the wheels, does not cause the car to want to rotate around it's "Center of Gravity", it simply causes the wheels to rotate around the axle
The force vector of the COG (with respect to gravity) is perpendicular to the ground not parallel, and it shifts towards the rear, as the front end lifts, due to the torque that the pinion applies to the ring in the differential.Originally posted by wimms
(SNIP) Seems you aren't relativizing enough. From your post it seems you imply that chassis lift occurs around rear axle. But you can't ignore inertia of car that is the only reaction force, concentrated to COG, which is located above rear axle and in front of it. Resultant force vector is parallel to the ground[/color], any other idea would violate some conservation law. COG cannot depart from ground, as this would be liftoff of the whole mass of the car and reduction of traction. The only possibility that remains is force vector directed ahead between COG and ground, causing chassis rotation around its COG. (SNoP)
NO!, but it is notable that forces acting in the two different "spaces", will offer different intertial results.Originally posted by wimms
(SNIP) Second, are you claiming that inertial effects that occur in space do NOT occur in gravity?? (SNoP)
Its unfortunate that I placed this 'force vector' in the middle of talks about COG, but I meant acceleration force vector. COG vector doesn't shift anywhere, unless you imagine spacetime curvature changes or car geometry changes. COG is center of MASS in gravity.Originally posted by Mr. Robin Parsons
The force vector of the COG (with respect to gravity) is perpendicular to the ground not parallel, and it shifts towards the rear, as the front end lifts, due to the torque that the pinion applies to the ring in the differential.
Whats that? Do you agree or do you disagree? Are you saying now that in gravity, the results would be COMPLETELY different, and therefore example of inertial effect in free space is, as you put it, non sense?NO!, but it is notable that forces acting in the two different "spaces", will offer different intertial results.
Sorry, but that simply isn't true.Originally posted by wimms
Its unfortunate that I placed this 'force vector' in the middle of talks about COG, but I meant acceleration force vector. COG vector doesn't shift anywhere, unless you imagine spacetime curvature changes or car geometry changes. COG is center of MASS in gravity.
So your link deals mostly with "braking forces" and does NOT offer explanations of why/how accelerative forces work to produce similar, but NOT identical results. He took the 'simplicity route' to explain weight tranfers, Braking effects. He avoided explaining how the accelerative effects cause front end lift, other then inertial mass, as inertial mass alone will not succeed in lifting the front end off the ground as the force vector does not go UP, it goes parallel to the ground, in accelerative force.from; The Physics of Racing, Part 1: Weight Transfer, Brian Beckman, physicist and member of No Bucks Racing Club, P.O. Box 662
Burbank, CA 91503, ©Copyright 1991
The braking forces create a rotating tendency, or torque, about the CG.
AND[/color]
It is a fact of Nature, only fully explained by Albert Einstein, that gravitational forces act through the CG of an object, just like inertia. This fact can be explained at deeper levels, but such an explanation would take us too far off the subject of weight transfer.
And from part 2; Keeping Your Tires Stuck to the Ground[/color]
That is, we explained why braking shifts weight to the front of the car, accelerating shifts weight to the rear
I don't understand you here. COG is point. Gravity Vector from it is directed always perpendicular down to the ground. No matter in what position object you hold is, no matter what other forces act on it. Are you talking about rotated direction of G vector relative to object geometry, or are you talking about location of COG within object, as COG shift would imply?Originally posted by Mr. Robin Parsons
Sorry, but that simply isn't true.
Take a plank of wood/board, find it's COG, draw a force vector down to the ground, (perpendicular) lift one end, and NOTICE that the force vector has now changed (Rotated) POSITION, RELATIVE to the GROUND.
Not at all. The very start of explanation goes with words:So your link deals mostly with "braking forces" and does NOT offer explanations of why/how accelerative forces work to produce similar, but NOT identical results.
So, although his approach is indeed simplified, he makes it very clear that acceleration and braking are equal here.Let us continue analyzing braking. Weight transfer during accelerating and cornering are mere variations on the theme.
...
These equations can be used to calculate weight transfer during acceleration by treating acceleration force as negative braking force. If you have acceleration figures in gees, say from a G-analyst or other device, just multiply them by the weight of the car to get acceleration forces (Newton's second law!).
Funny thing about this is that in the end its true, as rear diff is the only source of torque for acceleration too, but no, I believe you too much overestimate contribution of this effect.It is the force vector of the torque on the ring/pinion set, that causes the lift to arise, pun intended!
Yeah, but if you'd try that with dragster, it'd simply blow up, without walking anywhere.If you were to lock the ring in place, you would see the pinion "walking" itself around the ring, pulling whatever it was attached to, with it. (given all forces being what are needed to do that!)
But you are completely ignoring the fact that rocket exhaust is directed away from box at height of COG, not at ground height! Infact, it is more probable that acceleration vector is applied slightly ABOVE COG and slightly at angle towards ground to provide downforce. Same with that car that broke sound barrier on salt lake, its thrust was applied at or above COG.Although it is attached to the tracks (if I remember that properly) that is to prevent take off at speed (as that might cause lift) but when the force vector forward, is maintained parallel to the ground, there is no lift, even though there is a tremendous G force being applied. Inertia alone doesn't tend to lift the front end of the sled, not in/with the same force/manner that the rail lifts, as it is absent of what the rail has, torque exerted upon the axle to the frame/chassis.
Perhaps you should recall what rotational inertia is.As for spatial examples, (in absence of gravity) if your rail was in free space, and you started the engine, the torque from that alone, would initiate lateral rotation around the COG. Engaging the drive-train would probably initiate rotation of the frame/chassis around the axle, same as a helicopter without it's counterballancing rear rotor blades.
Now we almost agree. I'd only like to clear up what effect dominates, inertial rotation around COG as I understand, or ring/pinion torque to lift full length of chassis, as you say.Is that clearer? better? do we now agree?
Cheese I had thought you would get that one, cause it is simple.Originally posted by wimms
I don't understand you here. COG is point. Gravity Vector from it is directed always perpendicular down to the ground. No matter in what position object you hold is, no matter what other forces act on it. Are you talking about rotated direction of G vector relative to object geometry, or are you talking about location of COG within object, as COG shift would imply?
Force vector doesn't change, or I'd build antigravity device.
Yes he does, but he is not dealing with forces strong enough to lift one end of the car from the ground, as in with rails the accelrative force available exceeds the braking forces that the car can generate.Originally posted by wimms
So, although his approach is indeed simplified, he makes it very clear that acceleration and braking are equal here.
Just that that is the contibuting factor in actually getting the front end off of the ground, forward acceleration alone isn't enough.Originally posted by wimms
I believe you too much overestimate contribution of this effect.
If you bolted the tires to the ground, the machine would attempt to do an "a$$ over tea-kettle" flip, using the rear axle as the point of rotation.Originally posted by wimms
Yeah, but if you'd try that with dragster, it'd simply blow up, without walking anywhere.
MRP, you have all the pieces, but you aren't putting them together. COG is the place on the board where the string is tied, the string itself is the force vector. The string is NOT the COG. Clearly the place the string is tied to remains in the same place no matter how you orient the board (as implied by your post).Originally posted by Mr. Robin Parsons
Cheese I had thought you would get that one, cause it is simple.
Take a board, find the COG, tie a string to it, let the string hang down by gravitational activity, and let the string represent effective center of gravity (means where the COG acts).
Lift one end of the board, and note that, the string remains perpendicular to the ground, hence effective COG has shifted towards the lower end of your board.
(the point on the ground that COG acts upon)
Besides being wrong on its own, wimms said "moment of[/i]intertia," not just "inertia." Big difference. http://dictionary.reference.com/search?q=moment of inertianot enough forward motion for inertia to have worked
Picky, picky. Yeah, I know COG and COM aren't exactly the same thing, but I'm not driving my car in space. On the surface of the Earth in a car, they are at the same point.Originally posted by wimms
Imagine this car in space...
By torque, you mean accelerative force, not angular acceleration of rear wheel, right?
Yes. What lifts the car is best modeled as the force between the ground and the two back wheels. The actual component of the torque that goes to lifting the car depends on the mass and moment of inertia of the car: some of the force pushes the car forward and some of it lifts the front end. I think we're on the same page about this.By torque, you mean accelerative force, not angular acceleration of rear wheel, right?
Ummm, russ...Originally posted by russ_watters
MRP, you have all the pieces, but you aren't putting them together. COG is the place on the board where the string is tied, the string itself is the force vector. The string is NOT the COG. Clearly the place the string is tied to remains in the same place no matter how you orient the board (as implied by your post).
MRP, this is a simple definition issue. There really isn't any room to argue: The force vector is not the COG. The force vector acts FROM the cog. http://dictionary.reference.com/sea...%20of%20gravity
I understand what you mean, but you are saying it wrong. What you mean to say is that the angle of the force vector from the COG with respect to the car changes as it rotates when the front wheels lift .
The point on the ground does move, but it doesn't have a name. I'm thinking you decided to call it "effective cog" because you now realize you were wrong but don't want to admit it. This really isn't important enough for that. Let it go.
Butt out!Originally posted by wimms
(SNIP) COG vector doesn't shift anywhere (SNoP)
cause "relative to the ground" the COG does shift, towards the rear, the force vector motion/movement, towards the rear, PROVES that.Originally posted by Moi
(SNIP) So when I had mentioned the COG "shifting" I had meant 'relative to the ground' cause it does. (SNoP)
I'm here correcting misconceptions and misunderstandings. The two quotes you provided are one of each.Originally posted by Mr. Robin Parsons
Butt out!
Cause apparently you missed this one from page three...
cause "relative to the ground" the COG does shift, towards the rear, the force vector motion/movement, towards the rear, PROVES that.
The COG is a point, not a vector.PS Russ, from the start line, to the finish line, the real, and actual, 'COG' moves towards the rear of the car(!), without question!
@ 45 degrees (much simpler) it, the COG, remains exactly where it was when the stick was level, BUT in the interaction of the COG with gravity, and the vector from the COG to the ground, that shifts towards the end that remains on the ground, ~9" from it.Originally posted by russ_watters
(SNIP) But anyway, I'll bite: When a yardstick is horizontal, the COG is at 18" on the yardstick. So tell me then: when you lift one end 30 degrees above horizontal: where EXACTLY does the cog shift to? Express the answer as inches from either end (specify) of the yardstick. Providing the equation for figuring out the COG would be helpful as well. (SNoP)
Thank you. Finally.Originally posted by Mr. Robin Parsons
the COG, remains exactly where it was when the stick was level
Huh? Where exactly does it go?But I'll still tell you, when a rail, or a funny car, (for that matter) goes from the start line, to the finish line, the real, and actual 'COG' (THE POINT, Not the vector) shifts towards the rear of the car. No question! No doubt!
Yeah, I probably should have let it go, but this kind of thing gets on my nerves. I hate breaking out resumes because no-one no matter how smart knows everything. COG isn't the only thing you got wrong, MRP. Moment of inertia is another one and its not a high school level concept. Its not a big deal, but this thread is sophomore Engineering Statics and Engineering Dynamics. It looks to me from wimm's posts that he has taken these courses.Oh yes, BTW russ, I learned aout COG's in High school physics classes that occurred when I was in grade 10. That was waaaaaaay back when I was ~15/16 years old, or ~1971/72, OOOOOooops that's right, you were not even born yet!
Sorry russ, but no confusion, no error, no mistake, no question, no doubt, the COG (the point, not the vector through it} shifts towards the rear of the car/rail between the start line, and the finish line.Originally posted by russ_watters
Huh? Where exactly does it go?
I still think you're confusing different concepts here (static and dynamic forces). Again, the COG (if you don't believe me, read the definition) is a point on a body defined only by its geometry and gravitational interaction. The acceleration of the car has nothing to do with that.