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Where do weight transfer/acceleration forces go? 1

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Fabrico

Automotive
Aug 15, 2005
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Without consideration of aerodynamics, does a wheel driven vehicle become “heavier” against the ground while accelerating? Can a 3,000 pound car put more than 3,000 pounds of downward force against the pavement? How much downward force can be produced?
 
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Ok. There is added force pushing downward on the dummies head (you left out the mustache:eek:)), the seat springs are compressing, and so is the tire. There is even added force where the bottom of the rim sits on the tire. But no added downward force on the ground. It's amazing that any simple tire can split these two "distinct" forces with such precision!

If the unicycle was doing the exact same thing on a banked purpendicular surface, there would be 100% of it's weight plus the lateral, which has now become centrifigul, force. But you tilt the ground back just a few degrees and whala! The forces have been completely separated again.

I think not.

 
[ But you tilt the ground back just a few degrees and whala! The forces have been completely separated again. ]


if you look at it that way, then in order for your unicyclist to NOT slide off the track into the trees, he will have to lower the total input force of the tire.
 

In corners you need to deal with centripetal and centrifugal acceleration . . . those are acceleration vectors normal (perpendicular) and tangential (in the same line as) to the direction of your travel. And yes, these are lateral forces. Remember the experiment of spinning with a bucket of water? These forces are also employed by figure skaters to spin faster or slower.

One other simple issue: Newton did not say that F = ma.

He said that the force was equal to the time rate of change of momentum (mv). Simplified, F = d(mv)/dt. When the chain rule is applied and mass is constant, then F = ma. For a car, mass can be considered constant, but in a curve, force vectors are no longer rectilinear, and acceleration becomes a complicated ODE (dv/dt).

Civil Engineers understand this in making either wide, sweeping turns or banking tighter turns.

Thus, the lateral or centrifugal acceleration is actually pulling you towards the center of the turn, so you don't go flying off the curve as long as you are under the acceleration that balances the vectors.


 
That's a good explanation of the lean. Now, how does that relate to pressure placed against the ground? Or from another perspective, the ground's pressure against the tire? Are things that complicated for a unicycle?

An upward accelerating helicopter can have a pull on its rotor shaft that is much greater than the helicopter’s total static weight. It basically only pulls one direction. Why is a unicycle, car, or anything else, so different?

Virtually every force within the unicycle is on a single line with a single contact patch. How can it apply two distinct pressures going in two different directions against the ground?

I guess the question is: are lateral and vertical forces separated for teaching and analysis, or are they really separate?


 
Courtesy of Miss Barnes, my high school latin teacher:

centripetal = seeking the center

centrifugal = fleeing the center

(meaning colinear, but opposite in sense)
 
With regards to braking leads to oversteer, accel leads to understeer. Provided you're not locking the brakes or spinning the tires, this is true. If turning in a circle on a skidpad near the limit, an application of braking will tighten up the car (ie make it turn in more), just like letting off the gas in the middle of a long sweeper will unload the rear end and likely send you into oversteer. Take any fwd car and try to accel out of a corner and you'll get undeniable horrible understeer - from the combination of unloading the front and using up the available traction by longitudinal accel. Take any RWD car, that's understeering already, an application of gas will worsen the situation.

With regards to the combination of horizontal and vertical normal forces. I think that's why a negatively dynamically cambered tire, even more so than necessary to keep the contact area flat, provides more traction, since you're taking advantage of the horizontal component force to add to the compressing of the local points of contact. Thereby increasing the dynamic "coefficient of friction." True vertical load however will always sum to the weight, unless as previously discussed, c.g. is accel'ing up or deccel'ing down.
 
"With regards to braking leads to oversteer, accel leads to understeer. Provided you're not locking the brakes or spinning the tires, this is true"

Except when it isn't.

For instance, if you have forward brake bias, then braking in a turn will typically cause understeer.

"Take any RWD car, that's understeering already, an application of gas will worsen the situation."

Wrong. A gentle poke of the throttle on the production car I regularly drive will bring the back end out just a touch, very handy when cornering on gravel roads.

Thank you for your contribution.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
 
Greg,

You're talking about situations where the tires are way beyond optimum slip angles. For subtle, mid corner adjustments on the track, if you want to tighten up just a little, you let off or apply just a tad of braking, to lift the rear end induce slight rotation.

Pilun
 
Leave them, buddy. They live in an alternative world.
I don't know what kind of cars they are working with, but those are far from real on-road racing.
 
mhejjas

Play nice and objectively look through your arguments.

The original post makes no mention of race cars. This point is not mentioned until about the 30th post where you introduce it when confronted with some facts that contradict you statements.

Your statements by admission only apply to very specific circumstances introduced to the argument by you.

You only introduced these very special circumstances when your argument was contradicted, you then try to insist that the circumstances you introduced are the only parameters to consider and overrule the OP.

If you focus more on building knowledge rather than defending face, we might all learn and progress.

As far as automotive industry goes, I am only a hot rodder with I believe some flare for mechanical design. I have been involved in building and tuning a number of very successful race cars and boats.

Greg is a professional automotive engineer, currently employed by a very large automotive company that builds some very good handling large everyday passenger sedans but has a history in suspension design with some of the industries icons re excellent handling road going sports cars, sporting large and luxury sedans and race cars. I would not dismiss his view without considerable thought.

Regards

eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
I described two examples where a normal production car will experience exactly the opposite effect of your previous predictions.

They are not in extreme conditions, they are in the linear handling range.

However, to be fair " For subtle, mid corner adjustments on the track, if you want to tighten up just a little, you let off or apply just a tad of braking, to lift the rear end induce slight rotation"

Is fine, and typical behaviour.

But I have driven rear wheel cars with throttle-on understeer, and some with throttle off-understeer, and some with throttle on oversteer and some with throttle off oversteer (albeit probably nearer the limit). Some cars do both. I have driven front wheel drive cars with throttle on understeer (most Audis), and some with throttle off understeer (VW Golf Mk I), some cars do both. I have driven front wheel drive cars with throttle off oversteer (Mini).

So, I'd be very wary of generalizing.

With brakes I have certainly had brake induced spins, but most modern cars are set up to give brake induced understeer.

But on the car I drive most often, a slight addition of throttle, even when not driving hard, will move the tail out when cornering on a gravel road.


Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
 

Since you woke things up...
With regards to the combination of horizontal and vertical normal forces…True vertical load however will always sum to the weight, unless as previously discussed, c.g. is accel'ing up or deccel'ing down.

Questions still remain on the simple act of going straigt ahead.

On lateral and vertical forces, the “boss” says: “the forces are one.”

This means there is only one dynamic force acting upon the ground at whatever angle. If the front wheel(s) of the vehicle is/are off the ground while traveling forward, then there is only one force acting on one wheel of the 2 wheeled vehicle, or pair of wheels of the 4 wheeled vehicle.

The "boss" also said: "moments of inertia will be irrelevant in steady state cornering"

This of course would apply to steady state acceleration in any direction on a vehicle with any number of wheels.

Again, during steady state acceleration, the two above conditions are constant.

For clarity; some 2 or 4 wheeled race vehicles can initiate and maintain a wheelie without involving inertia. With pliable suspension, the same type of vehicle can wheelie without raising the height of CG above the ground. Obviously this represents a wheelie induced by torque alone. Proof of these conditions are shown in low level wheelies maintained for several hundred feet, accompanied with no letup of the throttle or rate of acceleration. If CG becomes too high or unstable, for whatever reason, acceleration will cease. A wheelie from inertia or balance cannot represent steady state acceleration. How do all the wheelies relate to you and me? All vehicle acceleration falls somewhere between sitting still and maintaining a low level wheelie.

All vehicle acceleration involves weight transfer. In a straight line, all acceleration is steady state or more. Tire contact patch is no longer perpendicular to the chassis centerline. CG is not relavant.

Questions remain:
What is the angle of the one non-vertical force?
What is the magnitude of the force?


 
Fabrico (Automotive) 23 Mar 06 18:50
Ok. There is added force pushing downward on the dummies head (you left out the mustache), the seat springs are compressing, and so is the tire. There is even added force where the bottom of the rim sits on the tire. But no added downward force on the ground. It's amazing that any simple tire can split these two "distinct" forces with such precision!

Virtually every force within the unicycle is on a single line with a single contact patch. How can it apply two distinct pressures going in two different directions against the ground?
I guess the question is: are lateral and vertical forces separated for teaching and analysis, or are they really separate?

----> Ok on the topic of the unicycle. No, the lateral and longitudinal are not seperated and applied at 2 different magical tyre contact patches. For analysis it is split into lateral and longitudinal co-ordinate system relative to the ground so to make calculation of the actual force witnessed by the centre of mass of the item easier.

As you said earlier the neck witnesses a higher compressive force. This is perfectly true. But this is because the neck axis isnt in the vertical plane anymore, as its on the angle the neck also sees a fraction of the lateral force. So the ground itself does see more load from the tyre, BUT its exactly the same force in the vertical direction.


 
As you said earlier the neck witnesses a higher compressive force. This is perfectly true.

How can the neck witnesses a higher compressive force, but not other components such as the butt, axle, wheel, or tire?

But this is because the neck axis isnt in the vertical plane anymore, as its on the angle the neck also sees a fraction of the lateral force. So the ground itself does see more load from the tyre, BUT its exactly the same force in the vertical direction

A fraction of the lateral force? There is only one force and it's fully experienced by the neck and everything else.

Perhaps I missed it if someone mentioned it, but this seems to point to something that may make both sides look a little more sensible.

As far as a unicycle, it seems that downward vertical force from weight must decrease as acceleration increases. In other words, an acceleration that requires the unicycle to hold a 45* tilt may apply less, perhaps half, the vertical pressure from weight it had at rest. No?

While vertical pressure from weight goes down, vertical pressure from G force goes up. It's balanced and is neither fully vertical or horizontal. If you accelerate in a big bowel, the entire force would eventually become "lateral".

Of course this leads to questions about accelerating past that 45 degree mark, and adding long weighted arms to the mix. Obviously the G-force can become considerable, multiples of the original weight.

 
Fabrico, you and shenanigans are both right. You consider "downforce" to be the combined force that the rider feels in his downward direction. Others consider downforce to be the vertical component in an outside frame of reference. If a unicyclist was riding around on a giant bathroom scale the reading would never change (assuming the c.g. never accel'ed up or down). But the rider would definitely feel more "downforce" from his frame of reference.
 
I idnt read all of the commets so i may repeat some.

If we set the dragrace 4-linkage systm up in such way that it separate the rear axle from the chassis and thereby creating a lifting force at the chassis, we should for a period of time exceed the total car weight at the rear wheel. Front wheel is of ground slightley. The duration may be adjusted by shock absorber setting.
As the moost important for et times is the first feet of acceleration, we can pay the price of less rear wheel weight as the car retain to its "normal" height longer up the strip.

Seen over the total strip the average weight on ground is the same, as I understand the situation.

Goran Malmberg
 
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