noob question about ride height
- Thread starter MADTJ
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- 58
I think off-the-line acceleration might benefit from having the car already tilted back as much to start with, i.e. high front end and low rear end. Which would of course be detrimental to handling, and produce the opposite of ground effect once the wind picked up (50 - 60 mph?). Tilting back puts more weight on the rear tires. Put enough weight on the rear tires, as in putting the motor over the rear tires, could induce a quick back-flip, hence wheelie bars and the long wheelbase of rail cars to counteract this tendency.
- 370
i get it now, ive been thinking about the 30mm off the front thing duck and i think i understnad now. when u accel, the car pushes u forward, but the weight travels to the back. the 30mm would make it so u wont bottom out and that the chasiss will ush ur cars rear down causing more traction. am i rite? thats what im guesing is whats goin on.
djaft3rb3ats
djaft3rb3ats
Yes, it will give you more traction (which is a good thing), but your theory of raising the rear isn't correct. If you want to know what chassis downforce is, read this post by rk.djaft3rb3ats
If you don't understand it, then a no-brainer way to use chassis downforce is to raise the rear at least 30 mm more than the front for street cars. For race cars, you can read my post above.
- 24,120
- Midlantic Area
- GTP_Duke
Actually, the ride height has nothing to do with static weight distribution. You could jack the front end off the ground and it wouldn't put more load on the back wheels. So you want the rear high on a RWD car to let the chassis rock back under acceleration, which does transfer weight to the drive wheels.doormeister
- 370
thanks duck, no beef rite?
- 58
Duke
Sure it does. As you tilt the car back, it progressively puts more weight on the back tires until you reach a theoretical vertical body position with 100% of the weight (less the weight of wheels and tires) on the back tires.
- 466
Gawd, where do you people get some of this stuff. Ok, Einstein, let's experement, shall we? By the way, today is the 100 year anniversary of the publishing of the paper that touted the theory of relativity, maybe we can improve upon it. We are not counting weight transfer form acceleration, right? For the sake of experementation, let's say our virtual car weighs 600lbs, ok? And if no one minds, since it shouldn't affect the results, let's remove the wheels, shocks and springs and replace them with wooden posts, we will leave short square ones in back, but we will be switching out the posts up front. First test, all posts the same length. Put a bathroom scale under each post, what does it read? Right, 150lbs. Now, let's jack the front posts to a length where the chassis is raked backwards 45 degrees. Now by doormeisters calculations we would have what, 225lbs on the rear wheels and 75 on the front, or some funky ratio like that, huh? Wrong. Unless we have put weight BEHIND the back square posts, which we didn't; the weight distribution remains the same 150lbs/wheel. As you jack the car toward vertical, all weight remains evenly distributed, for the same reason that you can walk around on a raft without it flipping over. Once you reach vertical, in fact at the VERY INSTANT you reach vertical, all the vehicles weight will be resting squarely on the two back posts. If you were to continue past vertical by pushing it and there were no wheelie bars to take their HALF of the vehicles weight, it would tip over to again assume a position of equilibrium, rocking gently on the roof. If what you said were true, doormeister, we would have a lot more dump trucks tipping over backwards from too much load, how often does that happen?doormeister
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rk
It seems you just made my point, unless you are saying that going between 1/10th degree from vertical and pure vertical creates a change of weight on the front posts from 150 lbs to 0 lbs.(?) The transition follows a sine curve. But its been about 25 years, so I don't readily identify it as a cosine, sine, etc. The change would be minor in the first 30 degrees of tilt, but rather dramatic in the last 30 degrees, that is, going to a vertical postion. yea, I was very good at math. busted the curves in high school. 4 semesters of calculus in college. but that don't matter. I love this game, so I enjoy talking about it. peace.
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Two guys carry a piano up the stairs. They are the same size, same strength. Who carries the load? Does it make a difference if the stairs are shallow, like an incline or steep, like a ladder? Heck yes.
- 466
A sine curve is a mathematical function equal to the vertical coordinate of a circumference point divided by the radius of a circle with its center at the origin of a Cartesian coordinate system, which is a careful way of saying at any point on the sine curve the real number value of it's x co-ordinate is equal to the real number value of it's y co-ordinate, i.e.: if you tilted the car 50% the weight distribution would shift 50%.doormeister
That is incorrect. The transition actually graphs to a VERY steep parabola.
It has been only about 20 years for me.
Are you really trying to compare an inclined plane to what is basically a weight at rest? I can see you still don't understand.doormeister
Put a 50 pound weight on a chair. How close to the edge can you place it before the chair tips over? The answer, of course, is that the chair will not tip until you have placed enough of the weight BEYOND a line drawn between any two legs, "enough" being a little more than half of the 50lbs and dependant on the width of the chair, the weight of the chair and the length of its legs. Or think of a tripod. You can make any two legs almost vertical, indeed they can be 1/100th of one degree less than vertical and all three legs will bear the same weight, they will make equal dents if placed on level sand.
With your piano illustration, you have taken the car and placed it on a hill, that is a completely different environment than adding length to its legs on level ground.
It's ok, curve buster, there are several kinds of math, I met my match at differential calculus, but I am an idiot savant with anything less, wanna go another round?
.
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rk
I don't think placing the car on a hill is different than adding length to its legs on level ground. We can disagree.
I am talking static positioning here, which seems to be what we are discussing...
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If this is your tripod, no, the vertical legs will put a bigger dent in the sand. Imagine continueing to tilt the third leg to the horizon and eventually almost horizontal. if the leg has no weight itself, all the weight being from the camera or whatever is on the tripod, the third leg will carry no weight as it moves to the horizontal position.
well the graphics came out wrong when it posted. I had to put in periods to maintain the spacing of the dashes.
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If this is your tripod, no, the vertical legs will put a bigger dent in the sand. Imagine continueing to tilt the third leg to the horizon and eventually almost horizontal. if the leg has no weight itself, all the weight being from the camera or whatever is on the tripod, the third leg will carry no weight as it moves to the horizontal position.
well the graphics came out wrong when it posted. I had to put in periods to maintain the spacing of the dashes.
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- GTP_Duke
It's very different.doormeister
On flat ground, the axis of load is perpendicular to the ground, no matter how long the posts (ie, the length of the springs) are. As you make one end or the other taller, the center of gravity of the car remains the same and the posts move closer together (infinitessimally, in real terms, or dramatically, in our theoretical example), but the CG stays firmly planted halfway between the posts (assuming the ideal 50/50 weight distribution in our ideal car). But because the ground is flat, the load is always carried straight down, meaning the static weight distribution is NOT affected by the rake of the car. The static load distribution doesn't begin to change until the CG actually begins to fall over one set of posts (ie axles).
[edit] That's what's happening in your tripod analogy. [/edit]
But if you put the car on a hill, the force of gravity is no longer perpendicular to the ground plane. There's now a lateral vector to it. This is why the piano is heavier for the guy on the bottom. If you park a car on a hill its static weight distribution changes, too. But that's because of the angle of the ground relative to the force of gravity, not because of the angle of the car relative to the force of gravity.
- 466
A tripod, by its very nature, will fall into a even level position across the bottom of its feet. It doesn't matter if you have two legs in New York and a third in London. If there were no weight on its extended leg, what on Earth would keep it from falling over away from London?
When you have two men pick up a "tilted", piano the man who picks up the long legs holds as much weight as the man with the stubby legs. Really this becoming absurd. When the men walk up a staircase, they are no longer able to easily balance the piano, gravity pulls it downward. Since the man above has little purchase, he can do little more than help to balance it as the bulk of the weight shifts to the lower man who is to some extent UNDER the piano. If they had handles, they would be able to share the load equally, regardless of the incline, so long as it wasn't vertical.
If that doesn't work for you, think of a teeter-toter. Given equal mass at either end, it will ALWAYS be in a state of balance, regardless of the angle it sits at, otherwise it wouldn't work.
We can certainly differ in opinion, Newton (and Duke) would agree only with me. I don't even care if you are convinced, I am concerned that you do not confuse others.
Tilt a car up to put more static weight on a set of wheels and you will see me do this:
edit: I wrote this last entry before I saw Duke's most recent post (I am also cooking dinner for my son). I am sure he would agree with my assertions up until this last post, he may not agree entirely with the assertions in THIS post.
When you have two men pick up a "tilted", piano the man who picks up the long legs holds as much weight as the man with the stubby legs. Really this becoming absurd. When the men walk up a staircase, they are no longer able to easily balance the piano, gravity pulls it downward. Since the man above has little purchase, he can do little more than help to balance it as the bulk of the weight shifts to the lower man who is to some extent UNDER the piano. If they had handles, they would be able to share the load equally, regardless of the incline, so long as it wasn't vertical.
If that doesn't work for you, think of a teeter-toter. Given equal mass at either end, it will ALWAYS be in a state of balance, regardless of the angle it sits at, otherwise it wouldn't work.
We can certainly differ in opinion, Newton (and Duke) would agree only with me. I don't even care if you are convinced, I am concerned that you do not confuse others.
Tilt a car up to put more static weight on a set of wheels and you will see me do this:
edit: I wrote this last entry before I saw Duke's most recent post (I am also cooking dinner for my son). I am sure he would agree with my assertions up until this last post, he may not agree entirely with the assertions in THIS post.
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Duke
So at what exact tilt does it change from being even distribution to a shifted distribution? 89 degrees from horizontal? 89.999 degrees? Please enlightned me. Please tell me at what exact position our scales would suddenly switch from 150 lbs per wheel to 300 lbs on each back wheel.
- 466
You DO know what a parabola is, don't you? We are discussing absolutes in a real world. The fact is that the 50 lb weight cannot occupy a single theoretical point in space, if it could, the parabola would be almost perfectly "square", it's "asymptotes" (this term actually describes the legs of a hyperbola, but hopefully we are not mincing semantics) would be nearly parallel, the weight would shift as the singularity crossed over a line drawn between the two legs. Since the weight must have volume, the act of crossing the line would be less immediate, and the resulting parabolic graph of the weight transfer would be rounder.rk
You can use the above example to answer your own question. If you can't, I will gladly break it down into bite sized chunks for you.doormeister
- 58
I've had a few minutes break from this, and I think my point in my very first post should have been made with a center-of-gravity arguement.
(The chair tips once the center of gravity crosses to outside the four legs (any two legs). If the center of gravity is .001" outside the 4 legs, it can tip. simple enough.)
I now believe that each wheel can measure 150lbs even as the car is tipped into a vertical position. Doing so, with the post example, would mean 4 posts in a line (say, if the front wheels/posts were wider than the back wheel/posts, and could line up). I know I can distribute weight evenly accross 4 posts in a line, hence 150lbs each. This is contrary to what I stated earlier, and in no small way. (You feel better?)
What I had in mind originally was basically what Duke said: as you tilt the car back, the center of gravity moves closer to being over the rear wheels. Granted, this would be a very small amount in the range that we are using in the game. We seek this for traction in the rotation of the body under acceleration, and can influence it by initial setup. If you have a long car, with the engine in front, the center of gravity is way out front and does not help your traction much even under acceleration. But if the center of gravity can be moved to right in front of the rear axel, as with a rail car, then that can be used to your benefit under acceleration. So I should have originally said that you can move the center of gravity closer to the rear wheels by lowering the back and raising the front. I hope this is not something you can argue with, since it is stated clearly in Duke's post...
I will make a note that RK thinks I am an idiot...
(The chair tips once the center of gravity crosses to outside the four legs (any two legs). If the center of gravity is .001" outside the 4 legs, it can tip. simple enough.)
I now believe that each wheel can measure 150lbs even as the car is tipped into a vertical position. Doing so, with the post example, would mean 4 posts in a line (say, if the front wheels/posts were wider than the back wheel/posts, and could line up). I know I can distribute weight evenly accross 4 posts in a line, hence 150lbs each. This is contrary to what I stated earlier, and in no small way. (You feel better?)
What I had in mind originally was basically what Duke said: as you tilt the car back, the center of gravity moves closer to being over the rear wheels. Granted, this would be a very small amount in the range that we are using in the game. We seek this for traction in the rotation of the body under acceleration, and can influence it by initial setup. If you have a long car, with the engine in front, the center of gravity is way out front and does not help your traction much even under acceleration. But if the center of gravity can be moved to right in front of the rear axel, as with a rail car, then that can be used to your benefit under acceleration. So I should have originally said that you can move the center of gravity closer to the rear wheels by lowering the back and raising the front. I hope this is not something you can argue with, since it is stated clearly in Duke's post...
I will make a note that RK thinks I am an idiot...
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I already told you. Once you're tilted far enough that the CG actually begins to fall over the contact patch of the rear tires. This is assuming the CG is at the center point between front and rear axles, as I've already said, giving ideal static weight distribution of 50/50.doormeister
Now, it's possible to jack the weight distribution around by cornerweighting the car, in real life. This calls for being able to set the ride height independently at each wheel, which we can't do in GT4.
But I didn't say that. Because as you tip the car, the wheelbase gets shorter (relative to the flat, level ground). So for every inch the CG moves closer to the rear axle, the wheelbase is actually getting shorter by 2 inches (due the angle of the chassis relative to the ground; think of the car's shadow if the sun was directly overhead).doormeister
Consequently your static weight distriburion remains 50/50 until the actual point of the CG begins to move over the area of the contact patch (creating the parabola that rk describes). You're getting there, though! Stay with us.
👍
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Duke
I didn't disagree with you. The CG does move tword the rear wheels, the wheel base does get shorter, and the CG does stay right between the two wheels. I'm not sure what I said in my last post that implied otherwise. Can't stay with you on this though. We certainly don't communicate well; it's just a game, and the house is empty, so I'm going to go play.
- 466
Whew. Well, if nothing else, this thread should serve to completely answer any question about static weight distribution as applies to our tuning. It may seem like a lot to wade through, but we have certainly taken many more pages at this site over subjects that only approach this thoroughness. Glad we are all on the same, um, post. 
- 24,120
- Midlantic Area
- GTP_Duke
My mistake. I misinterpreted this:doormeister
I took that to mean that you were saying that the center of gravity moved towards the back of the car... and I inferred that meant away from the front, meaning changing the static weight distribution. My mistake.doormeister
👍
That's what it's all about! Have fun.
- 58
I really wish a mechanical engineer had chimed in on this. I made a statement in my second post:
Quote:
Originally Posted by doormeister
Sure it does. As you tilt the car back, it progressively puts more weight on the back tires until you reach a theoretical vertical body position with 100% of the weight (less the weight of wheels and tires) on the back tires.
and was greeted with 'Gawd, where do you people get some of this stuff.' from RK, which is a rather insulting response, not that it matters. I never said that the progression was linear.
RK responds:
As you jack the car toward vertical, all weight remains evenly distributed, for the same reason that you can walk around on a raft without it flipping over. Once you reach vertical, in fact at the VERY INSTANT you reach vertical, all the vehicles weight will be resting squarely on the two back posts.
So I asked what 'VERY INSTANT" meant, asking if that was at 89.999 degrees or whatever. Then RK goes on to say later that this transition follows a parabola, implying I'm a moron in the process. 'VERY INSTANT', which I take to mean 'instantaneous', and 'parabola' imply contradictory concepts.
I believe (with just a few minutes of head scratching) the weight transition is proportional to 1- cosine, which does define a parabola of sorts as you transition from 0 to 90 degrees. But please don't argue this point, because I would like a mechanical engineer to answer this, and not have us go round on it.
Tripod: The shorter, more vertical leg will always carry more weight than the longer, less vertical leg. Is there an engineer out there? Please let a mech eng. answer this if it is not immediately intuitive.
Hill vs. raised spring height: The load on the axles is straight down, regardless of whether there is a hill underneath or a group of midgets holding the car axles up.
So I do not agree with RK:
Whew. Well, if nothing else, this thread should serve to completely answer any question about static weight distribution as applies to our tuning. It may seem like a lot to wade through, but we have certainly taken many more pages at this site over subjects that only approach this thoroughness. Glad we are all on the same, um, post.
I think this discussion was filled with mis-information, and poor and incorrect examples. I would have let it go, except for the agressive and somewhat personal attacks. It is a game. I want to have fun, but you two seem bent on getting the story straight, which is not bad in itself. If that is the case, then lets hear from a qualified mech. engineer (my minor in engineering is not pulling much authority here...)
To see if what I am saying has any validity AT ALL, try your favorite car in the 400m, first with the front at max. height and the rear at min. height, then switch - front at min. height and the rear at max height. I was able to shave .020 to .050 sec. off my times by raising the front and lowering the back. I know this is contrary to every post on this subject I have seen, but just try it. Don't flame me about it. Just see if it works.
I know RK and Duke will respond, hopefully in a civil manner, but I truly hope a mech engineer can respond to these issues.
Quote:
Originally Posted by doormeister
Sure it does. As you tilt the car back, it progressively puts more weight on the back tires until you reach a theoretical vertical body position with 100% of the weight (less the weight of wheels and tires) on the back tires.
and was greeted with 'Gawd, where do you people get some of this stuff.' from RK, which is a rather insulting response, not that it matters. I never said that the progression was linear.
RK responds:
As you jack the car toward vertical, all weight remains evenly distributed, for the same reason that you can walk around on a raft without it flipping over. Once you reach vertical, in fact at the VERY INSTANT you reach vertical, all the vehicles weight will be resting squarely on the two back posts.
So I asked what 'VERY INSTANT" meant, asking if that was at 89.999 degrees or whatever. Then RK goes on to say later that this transition follows a parabola, implying I'm a moron in the process. 'VERY INSTANT', which I take to mean 'instantaneous', and 'parabola' imply contradictory concepts.
I believe (with just a few minutes of head scratching) the weight transition is proportional to 1- cosine, which does define a parabola of sorts as you transition from 0 to 90 degrees. But please don't argue this point, because I would like a mechanical engineer to answer this, and not have us go round on it.
Tripod: The shorter, more vertical leg will always carry more weight than the longer, less vertical leg. Is there an engineer out there? Please let a mech eng. answer this if it is not immediately intuitive.
Hill vs. raised spring height: The load on the axles is straight down, regardless of whether there is a hill underneath or a group of midgets holding the car axles up.
So I do not agree with RK:
Whew. Well, if nothing else, this thread should serve to completely answer any question about static weight distribution as applies to our tuning. It may seem like a lot to wade through, but we have certainly taken many more pages at this site over subjects that only approach this thoroughness. Glad we are all on the same, um, post.
I think this discussion was filled with mis-information, and poor and incorrect examples. I would have let it go, except for the agressive and somewhat personal attacks. It is a game. I want to have fun, but you two seem bent on getting the story straight, which is not bad in itself. If that is the case, then lets hear from a qualified mech. engineer (my minor in engineering is not pulling much authority here...)
To see if what I am saying has any validity AT ALL, try your favorite car in the 400m, first with the front at max. height and the rear at min. height, then switch - front at min. height and the rear at max height. I was able to shave .020 to .050 sec. off my times by raising the front and lowering the back. I know this is contrary to every post on this subject I have seen, but just try it. Don't flame me about it. Just see if it works.
I know RK and Duke will respond, hopefully in a civil manner, but I truly hope a mech engineer can respond to these issues.
