Downforce 101?

I am not sure that a 2-wheeled vehicle has much in common with a 4-wheeled vehicle where cornering forces are concerned.

 

'When you are riding a bike of any sort you are told to lean into a turn. If you don't then you simply fall off no matter what other factors are involved.'quote

Actually 2 wheeled are 'counter steered' not by 'leaning'.

When you 'lean' you are actually 'counter steering' by leaning you transmit pressure that turns the handle bars in the opposite direction that you are leaning.

This one of the major reasons that motorcyclists hit cars that turn in front of
them (whether they were seen or not). The natural instinct is to 'turn away'
and on 2 wheels that turns you into the object if you don't counter steer.

In go-karts since your weight is in the center and your mass is considerable
part of the vehicle, leaning into the corner you do have a positive effect on
the speed you can take the corner. By keeping more rubber on the track.

None of this has anything to do with 'down force' which is an aerodynamics,
by the way.

stephen

 

First off, a motorcycle has virtually no aerodynamic downforce. What it does have in a turn, is centrifugal force by leaning down on its side into the turn, lowering the center of gravity closer to the ground, and closer to the inside of the turn. The harder the turn, the farther the bike is leaned down, the greater centrifugal force is generated. Enough that the suspension sags down under the increased weight. Its not aerodynamic downforce, its centrifugal downforce.

In a car, there is absolutely no way to make the car lower its center of gravity, or move the center of gravity in any direction. In a car that can generate aero downforce, it generally does not take effect until speeds above 50-70 MPH are encountered. At those speeds, when manuevering around a turn, the air speed on either side of the car will still be more or less equal, and regardless, the vacuum generated under the car will be generally equal at all points.

 

The physics of a cornering motorcycle and car are pretty different. However, the tires respond similarly in both cases.
A car with 0 lift does not at all react like a leaning bicycle in a turn. Not sure what the heck you mean when stating "when you are riding a bicycle in that the inside edge would sink and the outside would rise."
Dowforce on a car increases only the downward force on the tire, which allows the tire to sustain higher cornering and/or accelerating forces. Unlike actual mass inside the car, this extra force doesn't pull the vehicle toward the outside of the turn, which would detract from the tires grip.
I'm actually not even sure what kind of answer you're looking for here. Perhaps it would be wise for you to learn some physics?

 

Centripetal force is what you experience in a turn. This force is provided by the pavement pushing against the tires. Since this force is applied at pavement level, which is below the center of gravity for both bikes and cars, a moment is generated about the CG of the vehicle.

When you lean into a turn on a bike, all you are doing is using gravity to generate a counter acting moment against that so you don't roll over away from the curve.

When you are in a car, the situation is very similar. If you simplified the case to think of the outside 2 wheels on the car as that of a bike. The CG of car is offset from the outside 2 wheels towards the corner. So in essense the CG of the car IS leaning into the corner like what you do on a bike.

All this has noting to do with aero downforce. Read up on high school physics and you'll understand centripetal force is a function of velocity squared over the radius of the turn. Read up first year university physics on force vectors and you'll understand how gravitational force from leaning balances out the centripetal force.

 

By shifting your weight to the inside of the corner on a bike, you also reduce the amount you need to lean the bike for a given turning radius.

By shifting weight I mean like this:
Image Unavailable, Please Login

 

Two comments:

1. If you put some more realistic numbers on your numerical example (say 3g at 120 mph), the maximum possible velocity difference inside edge to outside edge is more like ~2%, and

2. (which other's have said in different ways) You're comparing an apple to an orange -- yes, the downforce might be shifted a few percent in the outboard direction on a car (and it would better if it was shifted in the other direction), but it's still far inboard of the most outboard tire-to-ground contact location. If the downforce on a car acted outboard of the most outboard tire-to-ground contact location, that would be as daft as leaning the wrong way on a motorcycle .

 

the only good reason that race cars use downforce is because Ff = mu(s) * N

namely, the force due to friction is equal to the static coefficient of friction between the road and tires multiplied by the normal force of the car.
(we'll assume its static, and not kinetic or a mix, although a slight amount of slip will give better handling that pure static, but thats a little advanced)
the normal force is the force the ground "pushes" back on the car, perpendicular to the road surface
(this is used, because on inclines the gravity force changes from vertical to some angle off the vertical)

so where do all the aerodynamics come in?

downforce uses a redirection (and pressure differential) of airflow to push the car to the ground (or suck it to the ground in some cases). this increases the normal force. more normal force, as you might have guessed means more friction

but why would we want friction? doesn't that slow the car?

yes and no. as you come to a corner. you need the friction of the tires to pull you through the turn. if you were on a frictionless surface, there would be no turning at, as the front tires could not provide a side force to initiate a turn. the amount of friction required going through a turn is given by F= (V^2) / r. where F is the centripetal force as described above, v is your velocity, and r is the radius of the turn. to allow a greater V through the turn, you can increase the radius, or increase the F. increasing the F is due to friction (so more normal force or more coefficient of friction, like using race tires over street tires) or increase the radius (this is why driving the track requires knowing the "line" the "line" usually winds through the track as to create the greatest radius. come in wide, cut close and come out wide. this gives a very large radius (while keeping a relatively shorter distance all the same while)

conversely, more downforce, slows the car on a straightaway, aerodynamically; and due to the normal force increases friction in wheel bearings and such

if we solve our cornering force equation for speed and input the friction we will get something like this:

V = sqrt( r * mu(s) * N)

increasing any of those quantities will increase the cornering speed by the 1/2 power. i've explained how each of the parts of a race car can increase those three quantities.

theres the basic theoretical physics explanation.

there is however, the roll of the car due to inertial forces, which transfers weight side to side on the tires, and suspension is used to counter or reduce this. im not a suspension guru, so i wont go there. maybe someone else could explain that better than I, but that is quite advanced relative to the original post

as for motorcycles, i would have to think about it for a bit, but the posters above give a pretty good description.

hope that helps, and doesnt make anyone more confused, i know i went through it pretty fast, just re read it

 

Here's a little starter help:
http://www.icbm.org/cmgallery/index.php?cat=11

 

I don't think the posts cleared it up for you. Your physics is not wrong just the application. you are thinking infinately wide car withdownforce mikes more DF at the outside than inside. Yes but a car is not infinately wide and the car is never at the center of a radius but the outside of one. So in a macro way the DF acts as if on straight line.

 

Try this the next time you are riding your bike. Pick a straight stretch and keeping the bike verticle, take your left hand off the handle bars, now keep it off and turn the handlebars to the right as if to make a right hand turn. The results will surprize you. Will you turn right or will you turn left? The correct answer is you will turn left.
A much different set of dynamics happen in a bike that will never happen in a car involving the leaning of the bike as a necessary part of its function.