Lunar Dragger

I spent most of my Sunday figuring out some descent rates.

I believe there was an error in my Lunar Dragger article in that area.

I had casually assumed a constant vertical fall acceleration of 1/6 g
during the period of
horizontal velocity deceleration.

This is incorrect.

Here is my new thinking on this subject:-

A vehicle in orbit has the downward gravity force exactly
counterbalalnced by the upward centrifugal force. The latter is a
function of the square of the horizontal velocity.

Hence the downward vertical acceleration starts as zero, and increase as
the square of the lateral delta-vee.

Perhaps somebody would be kind enough to check the following math :-

This subject is expressed by the following algebra:

The following conventions are used:

h = horizontal
v = vertical
V = velocity
A = acceleration
i = initial
f = final
t = time
T = tf
^2 = squared

Av = k * DeltaVh^2 , where k = 9.8 / (6 * Vhi^2 )

= k * ( Vhi - Vh(t))^2 = k * ( Vhi - (Ah * t ))^2

= k * (Vhi-Ah*t) * (Vhi-Ah*t)

= k * (Vhi^2-(2*Vhi*Ah*t)+(Ah^2 * t^2))

Vv(t) = Integral of Av(t)dt

= k * ((Ah^2*t^3)/3)-(Vhi*Ah*t^2)+(Vhi*t))

= k * Vhi * t * ( t - (2*Vhi/3))

Hv(t) = Integral of Vv(t)

= k * ((Ah^2 * T^4 /12)-(Vhi * Av * t^3 /3)+(Vhi * t^2 /2))

= k * (Vhi * t^2 * (0.5 - (Vhi/4)))

========

If the above algebra is correct, then we get the following examples:-

For t (deceleration time)

t(sec) Vvf (m/s) Hvf (metres) Vvf miles/hr
1 -1.0878 0.40887 -2.44755
5 -5.41722 10.22194 -12.18875
10 -10.78 40.8877 -24.255
20 -21.3422 163.551 -48.02
50 -51.7222 1022.194 -116.375
75 -75.5416 2299.9375 -169.96875
100 -98 4088.77 -220.5

So the final impact velocity is even less than in my Lunar Dragger
article, by
about a factor of 20.

Since most of the velocity build up is during the final 10% of the drop,
it is best to free fall most of the way, do a braking burn shortly
before impact, with airbags doing the final cushioning.

I have put the spreadsheet on my website as
http://members.aol.com/cfrjlr/space/moondrop.xls

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