I have seen a couple of these discussions pop up lately:
Past discussions:
There are a few white papers and calculators out for this is well. I wanted to take a more simplistic approach to this and just look at wheel spacing ratios.
Assumptions:
All wheels are loaded and spaced evenly
6 wheel no drop has 2/3 effective length of 4 wheel
6 wheel drop has 1/2 the effective length of 4 wheel
8 wheel 2 center drop has 1/3 the effective length of 4 wheel
It seems the KOP drive is as good of standard as any in FIRST, which has had a 1.8 width to length ratio over the years. The current no drop KOP wide configuration has around a 2.5 with the 2/3 effective length of the 6 wheels. The drive in a day (DIAD) from vex appears to be around a 2.3 width to length ratio, but they have a few patterns to choose from.
Is the 1.8 number an acceptable minimum for width to length wheel contact patches?
Should there be a maximum width to length ratio?
Does center drop matter with length? (the opposite wheel clearance is twice the drop distance)
I don’t think it’s accurate to assume that the center wheels in a 6wd center drop are equally loaded. The center of mass is usually much closer to the center than either side of the robot and usually the center wheels take substantially more load (tread wear makes this very apparent). This means that teams can run quite substantially longer than the optimal ratio would suggest on a 6wd drop (see 2056 in 2013).
Aside from that nitpick this is a well done analysis. Please label your graphs better though. It’s hard to interpret a graph and spreadsheet of two unitless values where one of which isn’t properly labeled.
I’m not really sure where your numbers are coming from; are you assuming a certain length, width, or frame perimeter as constant, and what are you varying to get the different categories?
The KoP chassis does have a drop center, and can be constructed to several different lengths, which as I understand it, would change your numbers considerably.
The bottom lines as far as I can see are that:
For most solid tire materials, the track to “effective length” must be significantly greater than 1.0 to turn (1.25 to 1.50 are good enough for most purposes, 1.8 is rather high).
The center drop must be enough that you can only have significant weight on two axles. Longer robots with softer (pneumatic or sponge) wheels on soft carpet will require more drop than a robot with a shorter wheelbase, solid wheels, or which drives on firm carpet.
Agreed, its really CoG dependent. This is more of a worst case scenario of centered weight on 4 of the 6 wheels.
The middle row is a square configuration for wheel contact of the outermost wheels (1.00 W/L for 4 wheels). The 4wd is the baseline, and the other wheel configurations are calculated based on the assumptions. They are unit less.
I’m not sure I’ve seen this term before, but the idea seems to be the effective aspect ratio of the wheels which contact the carpet. A 4 wheel “tank drive” vehicle with normal (non-holonomic) wheels which is longer than it is wide will be unable to rotate or turn while driving because the lateral friction of the wheel on the carpet can generate more torque resisting the turn than the rolling friction can in creating the turn. This lateral friction is known as “scrub”.
However, making all robots “wide” isn’t effective either, as they tend to fall over backwards in hard acceleration or forwards in hard braking. The “drop center” drive is designed to give the best of both situations; a long wheel base for purposes of acceleration, and a short wheel base for purposes of turning (because the wheels which do not contact the carpet don’t count, and wheels which carry less than their “fair share” of the weight count less than those which do when it comes to generating the scrub torque.
Yes. I’m not sure what “the assumptions” were to generate the rows, either.
