Runways on the Moon?

First let me point out that the Moon Miner's Manifesto is not a peer
reviewed journal, so the article I wrote is mostly my BOE calculations
which have not been reviewed.

I would welcome folks checking over my numbers. I discovered one goof
after I had submitted it to MMM (it is correctred on the web page),
and there might be others.

>
> > During 50 seconds decelerating to zero from 1,500 metres/sec the
> > spacecraft will traverse a horizontal distance of 37.5 kilometers.
>
> OK. How do you calculate this? I want to put it in a spreadsheet (I

I assumed constant deceleration, so I took the average velocity and
multiplied by the application time.

Average velocity = 1500 / 2 = 750 metres /sec

750 m/s * 50 sec = 37,500 metres = 37.5 kilometres.

==========

The deceleration is (1500 m/s)/50s = 30 m/s2

1g acceleration is approximately 9.8 metres per second per second (9.8
m/s2).

1g applied for one second results in a velocity change of 9.8 metres
per second.

For BOE (or mental) calculations I usually round 1g to 10 m/s2

The G force is (30m/s2)/(9.8m/s2) = 3.061g (I rounded it down to 3g in
the article).

Assuming constant acceleration, distance travelled = t*(t*a/2)

a = acceleration = 3.061g = 30 m/s2

t = duration of acceleration = 50 sec

50*30*50/2 = 37,500 metres, or 37.5 km.

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