>Take a look at my article on the "Lunar Dragger" :-

Arrrrggh... I had thought of posting almost the exact same idea last week.
(Although with zero depth consideration on my part. It was at the concept
of 'couldn't we drag a long tether or something and use that for
deceleration?'). Coming up with 'something new under the sun' is not
trivial...

Glad to see you've done some serious thought on it.

For reducing vertical velocity I had considered launching 'shock proof'
cargo at the moons surface (say, large blocks of a elements useful for
silicon manufacture, etc.). Thus we aren't wasting mass on fuel.

I was also thinking of using the 'launch mass at the moon' for horizontal
velocity reduction. (We fly low, and aim at mountains...)

For horizontal or vertical velocity reduction I would suggest two mass
launches simultaneously with the angles forming a V. (This way the mass
would not be directly below the launch vehicle for vertical deceleration,
and the masses could be aimed at 'peaks' on the moon for horizontal
deceleration).

Since the mass launcher would probably use less space than equivalent
rockets. And we get to recover all of our reaction mass (aside from that
expended to power the electromagnetic launcher and needed for reorienting
the craft..). We have almost no loss of useful mass.

Tom M.
TomM@...

Additional comments

I think that the dragger should be sacrificial, however, it should be of a
raw material that is rare needed for lunar industry. (Carbon)

Although you did the preliminary calculations for Kevlar, it would be nice
if we could get some calculations for carbon nanotubes (although km plus
lengths are probably some time off... ). This would give double benefits as
a sacrificial material, because carbon is one of the many elements that is
rare on the moon, and is one of the most necessary. I don't like the use of
steel, because iron can be extracted from the moon, and thus would be
largely wasted mass.

I had initially conceptualized a flail design to a lunar dragger. The
advantage (I would think) would be that a flail has a higher surface for the
same amount of mass. (however then you need a 'spreader bar', plus
headaches about tangles, and rotation problems, so probably isn't as
attractive as I'd initially thought...)

The jerkiness problem had occurred to me, but I wasn't being clever and
didn't even consider playing out line. Very nice.

Tom M.
TomM@...

>
> Although you did the preliminary calculations for Kevlar, it would be nice
> if we could get some calculations for carbon nanotubes (although km plus

Since carbon nanotubes are nto commerically available I see little value
in doing
that calculation. I chose Kevlar because it is commercially
available. I have
since learned that there are other even better materials available,
e.g. Spectra2.

> lengths are probably some time off... ). This would give double benefits as
> a sacrificial material, because carbon is one of the many elements that is

Kevlar and Spectra2 contain lots of carbon. But most of it will be
vaporized by
the heat, and it will be difficult to extract useful quantities from the
residue which will be spread over a swath tens of kilometres long.

> rare on the moon, and is one of the most necessary. I don't like the use of
> steel, because iron can be extracted from the moon, and thus would be
> largely wasted mass.
>

I am not sure I follow that argument.

Steel is inferior to Kevlar and Spectra2.

But it depends on the point of origin. Tethers could be made from
lunar iron
which is widely available, and could be launched towards Earth fairly
cheaply with a mass driver.

This would be much cheaper than launching polymer tethers from Earth.

CR.

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Charles Radley wrote

>Since carbon nanotubes are nto commerically available I see little value
>in doing that calculation. I chose Kevlar because it is commercially
>available. I have since learned that there are other even better
materials available,
>e.g. Spectra2.

Fair enough.

>Kevlar and Spectra2 contain lots of carbon.

Yeah that occurred to me about 1/2 a second after I pressed send .

>But most of it will be vaporized by the heat

Well we have to over engineer it for a margin of safety, so the margin of
safety should be available. Also there is the length of material that is
spanning the distance from the craft to the surface which would not be
abraded, and thus would be less likely to be vaporized. Also, we have all
of the material that is left on the spool that we use for playing the line
in and out.

This would still appear to be a fairly decent amount of carbon that we could
recover.

>, and it will be difficult to extract useful quantities from the
>residue which will be spread over a swath tens of kilometres long.

I wasn't suggesting recovery of the vaporized material, only that which
remains relatively whole.

>But it depends on the point of origin. Tethers could be made from
>lunar iron which is widely available, and could be launched towards Earth
fairly
>cheaply with a mass driver.

>This would be much cheaper than launching polymer tethers from Earth.

Yeah, for some reason I failed to consider that - doh!.

Tom M.
TomM@...

>
> >But most of it will be vaporized by the heat
>
> Well we have to over engineer it for a margin of safety, so the margin of
> safety should be available. Also there is the length of material that is

The margin of safety is there to account for transients in stress,
I have not given any consideration to the thermal energy ablation rate.

There might be little margin from that aspect.

> spanning the distance from the craft to the surface which would not be
> abraded, and thus would be less likely to be vaporized. Also, we have all
> of the material that is left on the spool that we use for playing the line
> in and out.
>
> This would still appear to be a fairly decent amount of carbon that we could
> recover.
>

Yes.

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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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