Longer track

[FatSebastian]

Can you say it another way ...?

If it helps, Stan voiced something similar over here a while ago.

[Stan Pope]

Umm ... the "cant" number (as in the definition you gave) doesn't tell whether the axle is tilted up or down!

[Brian Clark]

Sorry for misusing terms, what I meant was camber vs cant. Thanks again for the help guys.

[davet]

... It also probably helps to employ positive camber on the DFW, rather than negative...

Yes!!! Excellent reminder! That reduces the sliding distance markedly, and, therefore, the energy loss during the coast, since the Work = Force * Distance!

What does this mean? A severe negative camber keeps the bottom of the wheel against the very bottom of the rail where it' less likely to get upset by a poor joint.

I've been trying to gather info on long-track setup myself for a 56' track we race on next month. Seems consensus is as little steer as possible to keep from losing speed on the flat. Keep COM as aggressive as possible to get as much advantage coming off the ramp as possible. Offset the weight to the DFW side so you can keep the COM aggressive and still run less steer. On our car with a 5-9/16" wheelbase we had a .65 oz DFW weight at 11/16" COM and no wobble and came in first in Districts. When we offset the weight we can get a 5/8" COM with .65 DFW. This means I can keep my 3 1/2" of drift in 4 feet but still move our COM back 1/16" of an inch. Hope that makes sense. Won't know though until our race.

[Stan Pope]

Yes!!! Excellent reminder! That reduces the sliding distance markedly, and, therefore, the energy loss during the coast, since the Work = Force * Distance!

What does this mean? A severe negative camber keeps the bottom of the wheel against the very bottom of the rail where it' less likely to get upset by a poor joint.

The DFW "sliding" is a bit difficult to visualize, but let's try it anyway. I think that there are two parts: (1) the outer tread edge of the wheel sliding on the flat of the track and (2) the inside edge sliding against the side of the rail.

The first part is easy; toe-in produces, in your sample case 3-1/2" of sliding friction for every 4' that your car travels down the track. The Work (loss) is the product downward force (0.65 oz) in your case X the sliding friction between wheel and track coefficient X distance (3.5" X 63' /4' or about 54") If you are able to retain stability with 1/2" drift in 4' with, for example, 1 oz DFW weight, then the sliding friction reduces to 1 oz X sliding friction coefficient X (.5 X 63'/4' = about 8". Compare 0.65X54 to 1.0X8.

The second part is kinda hard to visualize. If the toe-in is zero, the inner edge of the wheel contacts the rail at the wheel's lowest point, and the wheel rolls against the rail with no sliding. Add some toe-in and the contact point raises. How much it raises depends on the camber AND toe angles. Takes lots of trig to calculate exactly how much sliding takes place. I'll leave that as an exercise for the more motivated and/or precocious readers.

I've been trying to gather info on long-track setup myself for a 56' track we race on next month. Seems consensus is as little steer as possible to keep from losing speed on the flat.

Yes, I think so.

Keep COM as aggressive as possible to get as much advantage coming off the ramp as possible.

Yes, but you have to compromise a bit.

Offset the weight to the DFW side so you can keep the COM aggressive and still run less steer. On our car with a 5-9/16" wheelbase we had a .65 oz DFW weight at 11/16" COM and no wobble and came in first in Districts. When we offset the weight we can get a 5/8" COM with .65 DFW. This means I can keep my 3 1/2" of drift in 4 feet but still move our COM back 1/16" of an inch. Hope that makes sense.

I don't understand how moving the CM sideways on the car changes the DFW weight (unless the other front wheel is carrying weight ... you are running a 3-wheeler, right?) For a 3-wheeler, my analysis shows that the weight just transfers between the two rear wheels.

Won't know though until our race.

[Speedster]

Regarding Stan's comment about shifting weight sideways. I have a test car I've been working with so it was easy for me to try it.
The car is presently set up with a 1.8 oz. tungsten block behind the rear axle and a 2.0 oz. tungsten block in front of the rear axle. I was able to shift the 2.0 oz. block .238 inches sideways.
Block shifted to Left - DFW .56 oz, Right rear wheel 1.70 oz, Left rear wheel 2.74 oz.
Block shifted to Right- DFW .56 oz, Right rear wheel 1.88 oz, Left rear wheel 2.56 oz.
Regarding an aggressive COM. Without going into any detail at this time, the car ran faster by .02 seconds over 6 runs when the 1.8 oz block was in the back which gave a less aggressive COM. This was over 30' of travel on a circular arc track. What I have found the most interesting so far is how the slightest adjustment on the DFW affects the speed of the car. If you have available to you the same type track you will be racing on you have a definite advantage.

[Stan Pope]

Regarding Stan's comment about shifting weight sideways. I have a test car I've been working with so it was easy for me to try it.
The car is presently set up with a 1.8 oz. tungsten block behind the rear axle and a 2.0 oz. tungsten block in front of the rear axle. I was able to shift the 2.0 oz. block .238 inches sideways.
Block shifted to Left - DFW .56 oz, Right rear wheel 1.70 oz, Left rear wheel 2.74 oz.
Block shifted to Right- DFW .56 oz, Right rear wheel 1.88 oz, Left rear wheel 2.56 oz.

So, your test and my analysis agree. That is encouraging!

Sometimes it is good to have theory and experiment disagree, though. That is when one realizes that there is more to be learned. And progress happens!

Regarding sensitivity to toe-in setting ... once you learn the "ballpark" of best toe-in and camber for the DFW, consider going with reduced bend and "drill in" part of the toe-in and camber. Smaller bend will make the adjustment less "touchy".

[Speedster]

Will do, Stan.

[Stan Pope]

...

The first part is easy; toe-in produces, in your sample case 3-1/2" of sliding friction for every 4' that your car travels down the track. The Work (loss) is the product downward force (0.65 oz) in your case X the sliding friction between wheel and track coefficient X distance (3.5" X 63' /4' or about 54") If you are able to retain stability with 1/2" drift in 4' with, for example, 1 oz DFW weight, then the sliding friction reduces to 1 oz X sliding friction coefficient X (.5 X 63'/4' = about 8". Compare 0.65X54 to 1.0X8.

Compare these sliding distances to that of an axle in a 0.098" bore: 63". The combined bore-axle loss results from nearly 5 ounces against a (usually) well lubricated (low friction coefficient) surface, so the heavy toe-in "first part" losses appear to exceed 13% of bore-axle losses. In other words, these toe-in losses are really worth considering in the trade-off analysis!

[Stan Pope]

...
The second part is kinda hard to visualize. If the toe-in is zero, the inner edge of the wheel contacts the rail at the wheel's lowest point, and the wheel rolls against the rail with no sliding. Add some toe-in and the contact point raises. How much it raises depends on the camber AND toe angles. Takes lots of trig to calculate exactly how much sliding takes place. I'll leave that as an exercise for the more motivated and/or precocious readers.

Hmmm ... no motivated/precocious readers??? How about if I prime the pump a bit?

Observe that with positive DFW camber, as the DFW toe-in is increased the point of contact for the wheel moves forward and up. The contact point can be identified by the angle around the wheel rim measured from the lowest point of the rim edge. Let's call that the "contact angle." The sliding distance for the edge of the wheel against the rail appears to be the distance of travel X sin (contact angle).

Note that there is a limit to the contact angle that is imposed by the height of the center rail, typically 1/4" or about 1/2 the wheel radius. That will limit the contact angle to about 45 degrees and limit the sin (contact angle) to about 0.7.

To compute the work (loss) from friction against the rail we also need to find out how hard the edge of the tread presses against the rail. That force, I think, will be equal to the force of friction of the tread against the flat ... as was developed a few posts back in this thread. It can't be more, I think. (This paragraph needs to be studied a bit more to make sure I'm not "double dipping" on the force.)

So, anyone fresh on 3D geometry can maybe write an equation for us that computes "contact angle" from toe angle and camber angle?

The sum total of this part of the thread seems to reinforce the idea that there is a trade-off between drift (toe angle) and DFW down-force that come into play with varying track length.