8 sided Drive Train footprint

[orangemoore]

I have a question. Have I made an error in my math or is the 8 sided Drive Train much larger than a normal rectangle Drive train?

If I'm not wrong that is such an advantage.

[Joe G.]

No error. Angled sides also increase maximum robot footprint.

However, a lot of octagonal drive designs do not fully take advantage of this footprint. Generally, the wheels should be arraigned in a hexagonal pattern to maximize the robot footprint.

[AdamHeard]

Small issues. The min bumper length is 8 inches.

The 5 inches of taper you have is hard to effectively utilize with wider track width. We found 3.5 much easier to utilize with our drive layout. Other drives will have different results.

These changes will only slightly reduce the size of your comparison.

We just tried this on our comp bot and the point you raise is one of the pros high on our list. The perimeter size rules make logically laid out octagon bases attractive.

[Joe G.]

Originally Posted by AdamHeard

Small issues. The min bumper length is 8 inches.

Not necissarily. Per 2014 rules, segments less than 8 inches are legal as long as the entire side is covered by bumper, and adjacent sides are also legally protected.

Originally Posted by 2014 Manual

[Oblarg]

The optimal area you can get with a given perimeter constraint is a given by a circle. A regular n-gon becomes a better approximation of a circle as n increases.

Admittedly, yours isn't a regular n-gon, but in light of this, yeah, you shouldn't be surprised that an 8-gon gives you more usable area than a 4-gon.

[cadandcookies]

Geometry is such a lovely thing, isn't it?

[orangemoore]

Thanks for all the replies so quickly.

I am glad I can do simple math. Now to make use of my discovery.

[AdamHeard]

Originally Posted by Joe G.

Not necissarily. Per 2014 rules, segments less than 8 inches are legal as long as the entire side is covered by bumper, and adjacent sides are also legally protected.

Thanks! I overlooked that.

We can further optimize our octagon shape in the next rev...

That adds a little bit!

[Oblarg]

Originally Posted by cadandcookies

Geometry is such a lovely thing, isn't it?

It can be, though it's easy to overlook its subtleties and a often pain to iron out the details.

(If you want to be convinced of this, try actually proving the seemingly-obvious thing I posted above).

[DampRobot]

Originally Posted by Oblarg

It can be, though it's easy to overlook its subtleties and a often pain to iron out the details.

(If you want to be convinced of this, try actually proving the seemingly-obvious thing I posted above).

Challenge accepted!

In a regular n-gon, the central angle for each side is 180/n degrees, and if the ngon has an inradius r, the side length of each side is S=2r*tan(180/n). I proceeded to construct a isosceles triangle where the center of the n gon is connected with equal length legs to one side of the n gon. The inradius of the n gon is the altitude of that triangle. Thus, r=S/(2tan(180/n)). This means the area of that triangle is A=rS/2=S^2/(4tan(180/n)). Therefore, the area of an N gon with side length S and side number n is A=nS^2/(4tan(180/n)).

The perimeter of a n gon is p=nS. Therefore, p/n=S. Thus, A=p^2/(4n*tan(180/n)).

With p held constant, the limit of A as n approaches infinity converges upwards to the Area of a circle with radius r.

Therefore, a circle is the optimum shape for an n gon of a given perimeter, and the more sides a regular n gon has, the more closely the it approaches the greatest possible area it can have for a given perimeter.

[Oblarg]

Originally Posted by DampRobot

Challenge accepted!

In a regular n-gon, the central angle for each side is 180/n degrees, and if the ngon has an inradius r, the side length of each side is S=2r*tan(180/n). I proceeded to construct a isosceles triangle where the center of the n gon is connected with equal length legs to one side of the n gon. The inradius of the n gon is the altitude of that triangle. Thus, r=S/(2tan(180/n)). This means the area of that triangle is A=rS/2=S^2/(4tan(180/n)). Therefore, the area of an N gon with side length S and side number n is A=nS^2/(4tan(180/n)).

The perimeter of a n gon is p=nS. Therefore, p/n=S. Thus, A=p^2/(4n*tan(180/n)).

With p held constant, the limit of A as n approaches infinity converges upwards to the Area of a circle with radius r.

Therefore, a circle is the optimum shape for an n gon of a given perimeter, and the more sides a regular n gon has, the more closely the it approaches the greatest possible area it can have for a given perimeter.

This is one portion of the correct proof, but a crucial (and rather more difficult) piece is missing:

This proves that a circle has a greater area than any regular n-gon, and that the area of a regular n-gon approaches that of a circle as n increases. It does not show that a circle has greater area than any n-gon, regular or irregular. So, to complete the proof, you must show that a regular n-gon has the greatest area of any n-gon with a fixed perimeter.

Actually, if we're going to be really precise, there's an additional step in showing that if a circle has greater area than any polygon with fixed perimeter, then there can be no shape with fixed perimeter with greater area than a circle, but this follows rather trivially from the fact that we can easily approximate any (convex, piecewise-smooth) shape to arbitrary precision with an inscribed polygon.

[BBray_T1296]

I would much rather stick to an octagon than bending 3/4" plywood around a 17.825353626292277606114981497722" inside radius




(since we are getting so precise that we say a circle is more useful than any reasonable N-gon, I figure you want to be spot on 112" perimeter)

[Oblarg]

Originally Posted by BBray_T1296

I would much rather stick to an octagon than bending 3/4" plywood around a 17.825353626292277606114981497722" inside radius




(since we are getting so precise that we say a circle is more useful than any reasonable N-gon, I figure you want to be spot on 112" perimeter)

If you find a supplier who'll sell you plywood with 32-sigfig tolerance on the sizing, let me know.

[BBray_T1296]

Originally Posted by Oblarg

If you find a supplier who'll sell you plywood with 32-sigfig tolerance on the sizing, let me know.

I know. I still wouldn't want to bend plywood to a 18" radius