Anyone knew that the person was stable in his/her that a car was a low center of gravity somehow, but we tried it to prove this for oneself logically.

And I had T who was awfully good at physics cooperate and finished the proof at last.

Because you make it with knowledge knowing as of a thing of a tenth grader, there may be mistakes, but approve it.

Figure 1

(proof) It is said that there are overall height Xcm, car high X/10cm, car width Xcm, a car of weight Mt. In addition, this car assumes it 1 ※ rigid body. When this car inclined to angle θ, I calculate tire and point of contact A of the ground how a moment hanging to the car as an axis changes by a center of gravity.

※In addition, the acceleration of gravity assumes it g, and the significant digit does not consider it.

When there is a center of gravity in the bottom of the body, the center (●) of the orange)

Angle blue = angle red = θ (I refer to figure 2)

Figure 2

①＝M, g = Mg (arrow yellow in figure 2)

②＝(1) ・ cosθ = Mgcosθ

③＝(1) ・ sinθ = Mgsinθ (arrow pink in figure 2)

Figure 3

Ⅰ＝X/2cm

Ⅱ＝X/10cm

(in I, II a length = power for each line)

Expression of the moment

+ direction…(3) X II = X/10, Mgsinθ…(4)

- Direction…(2) X I = X/2, Mgcosθ…(5) It is ,④ that a car overturns＞(5) therefore, X/10, Mgsinθ＞X/2, Mgcosθ1/5, sinθ＞At the time of cosθ

When I calculate this,

1/25, sin^2θ＞1-sin^2θsin^2θ＞25/26

sinθ＞5/ √ 26

sinθ＞0.98058

Than this

θ＞Approximately 78 degrees…(6)

When there is a center of gravity on the top of the body (I refer to purple ☆ figure 4)

Angle blue = angle green = θ (I refer to figure 4)

Figure 4

①'= M X g = Mg (in figure 4 an arrow of the indigo blue)<>br ②'=①' X cosθ = Mgcosθ (in figure 4 an arrow of the blue)

③Arrow) which is light blue in '=①' X sinθ = Mgsinθ( figure 4

In addition,

Ⅰ'= X/2cm

Ⅱ'= Xcm

Figure 5

Expression of the moment

+ direction…(3) 'X II = XMgsinθ…(4) '

- Direction…(2) 'X I = X/2Mgcosθ…(5) '

A car overturning here ④ '＞(5), time of ', is XMgsinθ＞X/2Mgcosθ2sinθ＞At the time of cosθ

When I calculate this

4sin^2θ＞1-sin^2θ

5sinθ＞1

sinθ＞1/5

θ＞Approximately 26 degrees…(7)

⑥The degree of the degree of leaning that time when a center of gravity is lower than ⑦ is permitted increases.

It may be said that the body is more stable in its low one than this in the center of gravity of the car.

The above is our proof.

Thank you for reading.