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I have wondered this because I have been figuring average scale speeds, does for instance a Porsche 962 IMSA body style differ from the 962C LH as far as aerodynamics and down pressure in scale apply if both have the same motor, gearing and tires???

Short answer: "No"

Longer answer: You would need to be running the cars under water to be able to measure the difference in downforce between the two bodies. Any differences you might feel on a track will be due to the weight distribution differences rather than any aerodynamic differences.

This should be good.

I like the short answer. On our track it doesn't mean anything BTW :)

Short answer: no.

Long answer: nnnooooooooo.

For home tracks you won't see any difference. On a big banked commercial track, you will. That
is why a particular body will become "the one to use". For example, in retro Can-Am, the Bryant
Ti-22 is the hot body.

Depends on the rotation of the earth at any given spot in relation to the wind!!! I think :text-yeahthat:

From everybody's experience the answer should be NO. But in testing at home my experience has been yes. However the differences could be do to overall weight and/or weight distribution instead of aerodynamics...but I still like to think wings make a difference at home.

For those of you who have a nice big oval track can you please do me a favour and conduct the following test. Select a car, any car and dial up the voltage so that the car circulates on the outer most lane as fast as it can without coming off and with the controller trigger fully depressed. Register 10 laps, noting the fastest lap, the slowest lap and the average of the 10 laps.

Now cut a 1" square of plastic or cardboard off a serial packet and tape it to the car so that it is mounted on the roof in a vertical manner and sitting so that it forms a large frontal area face on. Now repeat the 10 laps and note the times. How much did it change by.

Should be an interesting result.

Yep, I gave my silly answer earlier, but there is some aero effect. It did play a part when we were doing the land speed record runs in Cincinnati, but those cars were running faster than the typical 1:32 car by a good margin. Doubling the speed quadruples the aero effect -- cutting the speed in half reduces the aerodynamic effect by a factor of four. So where it was really just starting to play a part at land record speeds, divide that minimal effect by 4 when talking about a typical 1:32 car, flat-out, on a typical track where you can't even reach the car's top speed.

Digging up my aerodynamics from 45 years ago and purely by memory: drag force is given as:

F=1/2.K.A.Ro v^2

where

F: drag force (Newton)
K: drag coefficient (unitless)
A: projected area (sq m)
Ro: air density in kg/m cubed
v: speed in m/s

For our situation:

K: typical 1:1 car drag coefficients is around 0.3. Let’s accept our slot car is similar to that
A: assume the car’s width is 60mm, height is 40mm. Area is 0.06m x 0.04m = 0.0024m^2
Ro: density of air is 1kg/m^3
v: assume 20km/h. This is 5.55m/s

Plugging in all that gives me 0.01N. This is the force equivalent of like 1.1g of mass exerts downward in gravity.

I think we can safely disregard that…

I would be happy if anybody could identify any errors of principle or math: I last did this stuff many, many, many years ago….

:laughing-rolling:

I think I like the cardboard method of proving this better. Jim

F=1/2.K.A.Ro v^2

where

F: drag force (Newton)
K: drag coefficient (unitless)
A: projected area (sq m)
Ro: air density in kg/m cubed
v: speed in m/s

Well, unlike all you comedians, I'm taking this seriously. So, if I have it right, the more kilos of fig newtons being dragged at X velocity through an area of dense cubes, the higher the cooperation on the project. That's it, right?

Love that video!