Type "ODTS4"

Further in my pursuit of the most elegant chariot, I reconsidered Dr. Eppendahl's design and - thanks to the beauty of abstract diagrams - found, that one idler wheel could be eliminated simply by swapping the differentials. This may be difficult with LEGO blocks, but Fischertechnik provides units with identical dimension for the "red" and "green" configuration.

Equations:

(a:)	rot₂ + rot₅ - k * rot₃ = 0
(b:)	rot₆ + rot₈ + k * rot₇ = 0
(c:)	n₄ * rot₄ + n₁₁ * rot₁₁ = 0
(d:)	n₁₀ * rot₁₀ + n₁₁ * rot₁₁ = 0

The remaining constraints:

(e:)	rot₁ = rot₂
(f:)	rot₃ = rot₄
(g:)	rot₅ = rot₆
(h:)	rot₈ = rot₉
(i:)	rot₇ = rot₁₀

Composition:

(a,e=>k)	rot₅ = - rot₁+ k * rot₃
(b,g,h=>l)	rot₅ = - rot₉ - k * rot₇
(k,l=>m)	rot₁ - rot₉ = k * (rot₃ + rot₇)
(c,f=>n)	rot₃ = - rot₁₁ * (n₁₁ / n₄)
(d,i=>o)	rot₇ = - rot₁₁ * (n₁₁ / n₁₀)
(m,n,o=>p)	rot₁ - rot₉ = - rot₁₁ * k * n₁₁ * (n₄ + n₁₀) / (n₄ * n₁₀)
(p=>q)	rot₁₁ = (rot₉ - rot₁) * n₄ * n₁₀ / ( k * n₁₁ * (n₄ + n₁₀))

To make it "south-pointing" requires, that if rot₁ is held 0, rot₁₂ has to be exactly one turn when wheel 9 of diameter d₉ covered one full circle (centre at wheel 1's contact to the ground, radius equal to track width t):

(r:)	d₉ * rot₉ * PI = 2 * t * PI	d₁ * rot₁ * PI = 2 * t * PI
(s:)	rot₁ = 0	rot₉ = 0
(t:)	rot₁₁ = 1	rot₁₁ = -1
(q=>)	rot₉ = k * n₁₁ * (n₄ + n₁₀) / (n₄ * n₁₀)	rot₁ = k * n₁₁ * (n₄ + n₁₀) / (n₄ * n₁₀)
(r=>)	d₉ = 2 * t / rot₉	d₁ = 2 * t / rot₁
(n₄ = n₁₀) => d₁ = d₉ = t * n₄ / ( k * n₁₁)

Given Parameters (Your choice !)

Reference	Size	Description
4/10	n₁₀ = [teeth]	Spur gear, fixed to drive shaft 3/7 of differential
11	n₁₁ = [teeth]	Spur gear, fixed to pointer
k	k =	Differential reduction factor
T	t = [length units]	Track width

Derived Parameters

Reference	Size	Description
1/9	d₁ = [units of t]	Road wheel, free running on axle