Is Six Inches Worth Modeling?

A car spends less than 4 hundreths of a second in the zero-join transition. Though the ramp and flat join is abrupt, the trajectory over the join is just less than the distance between the axles of the car. The shortness of the car's experience on the join leads to the suspicion that its complexities might be avoidable; by ignoring it, more-or-less altogether. The arguement to win the day must be a geometric one based on the trajectory model.

Evaluate the model

The transitional trajectory is nearly a line at an abrupt join that is not steep. The slope of the line depends mostly on how far the center of mass is in front of the rear axle, CMx. Since L2 must be less than B, the distance between the axles, CMx/B gives us a fraction that acts as an index to the slope of the transition trajectory.

The slope angle, s, is nearly O(1-CMx/B).

For an extremely front-weighted car, CMx = B (CMy = 0), so s = 0 (zero).

For an extremely rear-weighted car, CMx = 0 (CMy = 0), so s = O (the ramp angle).

Just to illustrate, let CMy = 0 and * = 0, then

Forward weight, CMx = B: dh = Rr/cosO + Rftan(O/2)sinO - Rr which is nearly 0 (zero).

Forward weight, CMx = B: db = (Rftan(O/2))cosO + Rrtan(O/2) + B which is just over B

Rear weight, CMx = 0: dh = Rr/cosO + BsinO + Rftan(O/2)sinO - Rr which is about -BsinO

Rear weight, CMx = 0: db = (B + Rftan(O/2))cosO + Rrtan(O/2) which is about BcosO

For CMx = B/2, s = O/2. Using -21 degrees for O,

dh = Rr/cosO + BsinO/2 + Rftan(O/2)sinO - Rr which is about -B/6

db = (B/2 + Rftan(O/2))cosO + Rrtan(O/2) + B/2 which is about 19B/20

s = arctan(dh/db) = 9.95 degrees which is close to O/2, 21/2 = 10.5 degrees.

What if it is ignored?

If the transitional trajectory is "ignored" by letting L1 end at the join and L3 begin there, the dynamics will change little for cars weighted near the front and rear axles. For cars weighted more near the middle, the acceleration on the transition will be greater by about sinO/sin(O/2) which is almost two times.

Assuming a speed of 13 feet/second or 156 in/s, g = 386.088 in/s2, CMx = B/2, B = 6 inches and O = -21, the acceleration would be at most -gsinO = 138.362 inches per second per second through a distance of B/2.

Using the standard expression x = at2 + v0t, B/2 = -gsinOt2 + 156t gives an expression in t that we can solve for time.

0 = t2 - 156t/gsinO + B/2gsinO

t = 156/2gsinO +/- \[(156)2/4g2sin2O - B/2gsinO]

t = (78 +/- \[(12168-BgsinO)/2])/gsinO

t = (78 - \[(12168-(6)138.362)/2])/138.362 = 0.0196 s which is about 1/51 seconds

So the car's speed increases by about at = -gsinO(0.0196) = 2.71 inches per second.

Assuming a speed of 13 feet/second or 156 in/s, g = 386.088 in/s2, CMx = B/2, B = 6 inches and O = -10.5, the acceleration would be at most -gsinO = 70.359 inches per second per second through a distance somewhat less than B.

Using the standard expression x = at2 + v0t, B = -gsinOt2 + 156t gives an expression in t that we can solve for time.

0 = t2 - 156t/gsinO + B/gsinO

t = 156/2gsinO +/- \[(156)2/4g2sin2O - B/gsinO]

t = (78 +/- \[6084-BgsinO])/gsinO

t = (78 - \[6084-(6)70.359])/70.359 = 0.0392 s which is about 2/51 seconds

So the car's speed increases by about at = -gsinO(0.0392) = 2.75 inches per second.

Though one car spends twice the time accelerating on the transition, it's speed increase was still about the same as the other.