Quote:
Originally Posted by 114Klutz
I can explain. The physics supports this.
Lets take bot A and bot B, both bots have X amount of weight. The coefficient of friction between the wheels and the ground is Us
Bot A has 6 wheels, 4 of which are powered, so, 2/3rdsX * Us is the max tractional force provided.
Bot B has 4 wheels, 2 of which are powered, so 1/2X*Us is the max tractional force.
This is, assuming, of course, that the distribution of weight and the wheel spacing is perfectly even.
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Hmm....I think 114Klutz is on the right track, but slightly off in the explanation.
Bot A (1 wheel per axle)
____________________________
|back---------x------------front|
|___________________________|
/---\---------|----------/---\
(----)--------|---------(caster)
\ _ /---------|----------\ _ /
--------------|-----------------
/\----------|-----------/\
|----------|-----------|
|----------\/----------|
F1---------W----------F2
|<----L1---->|<----L2--->|
Sum of forces = zero: F1+F2=W
Sum of moments about point x = zero: F1*L1 = F2*L2
So, F1 = (W-F1)*L2/L1 = W*L2/L1 - F1*L2/L1
This simplifies to: F1 = W*L2/(L1+L2)
The max tractional force can be calculated by using: Ff=mu*F1
Or,
Ff = mu*W*L2/(L1+L2)
Bot A (2 wheels per axle)
__________________________
|back---x----------------front|
|_________________________|
/--\----|---------------/--\
(----)---|-------------(caster)
\ _ /---|---------------\ _ /
--------|---------------------
---/\---|----------------/\
---|----|----------------|
---|----\/---------------|
--F1---W---------------F2
|<-L1->|<------L2----->|
When the extra rear wheels are added, the center of mass (the W) shifts more towards the back (the picture is slightly exaggerated).
So, looking at the formula for traction force again:
Ff = mu*W*L2/(L1+L2)
L1+L2 is still the same but L2 has gotten bigger, which increases the traction force (Ff).
So, it makes sense that the robot with more wheels would push the robot with less wheels even though the weights are the same because the weight
distributions aren't perfectly equal.
114ManualLabor - maybe you could try powering all of the wheels - that way the distribution of weight doesn't matter as long as it is somewhere in between the wheel axles. I'd be really curious to see if the 2/axle bot could still beat the 1/axle bot. Then this would mean that mu does in fact change with surface area (to a very slight degree).