Apollo 10 contingency plans

[Luke Pemberton]

Δv = 311 * 9.80665 * LN((8273 - 350) / (8273 - 350 - 2631)) = 1,231 m/s

Rechecking my first calculation, I did not get exactly the same figures as Bob (I can't do math when it's gone midnight, not any more).

I used the same calculation, but I converted Allan's 80 kg to lbm, so arrived with 31 m/s gain in Δv. Accounting for suit and PLSS, a gain of 64 m/s (as per Bob). Again, it goes to show the design margins. Incredible stuff really.

[Bob B.]

I got exactly the same figures as Bob, a little under 30 m/s is gained in dV for a loss of 80 kg, unless we're both doing something daft. Maybe someone can check both our figures and put up the equations.

It's just a fluke that we got the same number - we were both off by about half but for different reasons.  You were off by half because an astronaut's mass doubles when he's in his suit and backpack.  I was off by half because I used kilograms when I should have used pounds.

[dougkeenan]

May I ask, why is the g-value being used related to the earth and not the moon? 

I think it's because the weight of the fuel used in the calculations is based on earth weight but would like to be sure.

[Luke Pemberton]

It's just a fluke that we got the same number - we were both off by about half but for different reasons.  You were off by half because an astronaut's mass doubles when he's in his suit and backpack.  I was off by half because I used kilograms when I should have used pounds.

I see that now. I used 80 kg and multiplied  by 2.2 to get lbm, which gave me 176 lbm. Your suit and PLSS in kg was close to my mass in lbm (being 204 kg). Complete fluke. 

It's too late for this now.

[Allan F]

The Earth G-value relates to the equation of rocket power. To get the exhaust velocity you take the specific impulse in seconds and multiply it with G in m/s^2 - and get the answer in m/s.

The number 311 in the above equation is the specific impulse of the LM's ascent engine.

[Bob B.]

About half its mass.  Δv is a function of the ratio of the fully fuelled mass to the empty mass.  If you cut the fuel mass in half, then you have to cut the empty mass in half in order the maintain the same ratio.

Actually it's a little less than half because a fully fuelled ascent stage had more Δv than it needed just to get into some sort of minimally acceptable orbit.  To get into a 9 nautical mile orbit, the LM needed a Δv of about 1,850 m/s.  Therefore, we can use this Δv to compute the exact amount of mass the ascent stage needed to lose. We have,

1850 = 311 * 9.80665 * LN((8273 - X) / (8273 - X - 2631))

where X is the require mass reduction.  Solving for X we get 2,488 lbm (1,128.5 kg).

 

[Luke Pemberton]

I think it's because the weight of the fuel used in the calculations is based on earth weight but would like to be sure.

Welcome to the board with your first post. It's a specific impulse and is a convenient geocentric quantity. The way I think about specific impulse is that it enables us to normalise the rocket thrust if we know the weight of propellants through the nozzle.

We have some aerospace engineers that write at the board, they will be able to explain it fully. This site helped me understand it a little:

http://exploration.grc.nasa.gov/education/rocket/specimp.html

[VQ]

Now that I think about it, were PLSS backpacks included on Apollo 10?

Probably, to retain the contingency EVA capability in case they were unable to hard dock after rendezvous.

[Bob B.]

May I ask, why is the g-value being used related to the earth and not the moon?

That's a question that I've seen frequently.  Although others have already explained, I'll try my hand at it as well.

The equation that we're using to calculate Δv is called the Tsiolkovsky rocket equation, typically written as

Δv = V_e * LN( m_o / m_f )

where V_e is the velocity of the expelled exhaust gases, m_o is the initial total mass, and m_f is the final total mass.  The difference between m_o and m_f is the mass of propellant burned.

In practice we don't actually use V_e, instead we use something called the effective exhaust gas velocity, denoted C.  (Explaining the difference between V_e and C is a complication that I'd rather not get into right now.)  Therefore, the rocket equation is more correctly written

Δv = C * LN( m_o / m_f )

The thrust of a rocket is given by

F = C * ṁ

where ṁ is the flow rate of the expelled mass.  Rearranging we have

C = F / ṁ

The term F/ṁ is a very useful parameter in describing the performance of a particular propulsion system.  In the SI system, F/ṁ has the units N-s/kg, while the equivalent in the imperial system is lb-s/slug.  Further note that the value of F/ṁ is different in each system of units.

Fortunately there is a simplification.  If we divide by g_o, standard gravity, then we obtain a parameter that has the units of seconds, as well as the same value in any system of units.  This new parameter is called specific impulse, and is given by the equation

I_sp = F / (ṁ * g_o)

where g_o equals 9.80665 m/s² in SI units.  We use standard gravity because 1 kilogram-force = 9.80665 Newtons by definition.

We can check units as follows, noting that 1 N = 1 kg-m/s²

(kg-m/s²) * (s/kg) * (s²/m) = s

Rearranging the I_sp equation we get

I_sp * g_o = F / ṁ

Therefore, by substitution

C = I_sp * g_o

Tsiolkovsky's rocket equation therefore becomes

Δv = I_sp * g_o * LN( m_o / m_f )

[Bob B.]

Now that I think about it, were PLSS backpacks included on Apollo 10?

Probably, to retain the contingency EVA capability in case they were unable to hard dock after rendezvous.

That would be my guess as well.

[Allan F]

Now that I think about it, were PLSS backpacks included on Apollo 10?

Probably, to retain the contingency EVA capability in case they were unable to hard dock after rendezvous.

That would be my guess as well.

On later missions, they ditched the PLSS before lunar takeoff, and retained the OPS to allow for EVA transfer - and also for EVA to retrieve film from the SM during trans-Earth coast.

[smartcooky]

May I ask, why is the g-value being used related to the earth and not the moon? 

I think it's because the weight of the fuel used in the calculations is based on earth weight but would like to be sure.

We've seen some very mathematical answers to this, however, I thought it was simply that fact the the issue isn't the weight of the fuel but the mass. If a rocket plus its fuel load is 1000kg, it might be "weightless" in orbit, but its mass is unchanged; still 1000 kg.

[Luther]

What if one of the astronauts had stayed on the surface, in a "I'm just going outside, I may be some time" manner? Would the ascent module have been able to make it then or would the reduced mass not make all that much difference?

Too little to make much of a difference.  Leaving behind an astronaut would gain less than 30 m/s, which is still far short of what would be needed.

Not enough, even if he pushed some ...

[Peter B]

Just out of interest, once the LM had landed and all immediate post-landing activities had been performed, was there any way to reactivate the Descent Stage and use it as a sort of Lunar Ascent Rocket First Stage?

I have vague memories of this as the basis of a hypothetical Apollo alternate-history story I came up with many years ago, when I knew only the tiniest fraction about Apollo compared with now.

And while I think about it, I suspect I may have been inspired to think about the idea by a radio play I heard part of a long time ago (late 70s or early 80s). The scenario of the play was a CMP going rogue and threatening to ignite the SPS while the CDR and LMP were on the Moon. As I didn't hear the whole play (IIRC I had a battery powered transistor radio under the blanket late at night and the batteries died) I didn't find out what happened. If anyone knows anything about the play I'd be curious to learn more.

[Allan F]

The descent stage was safed and deactivated by venting the helium using to pressurize the tanks. Once that is done, the engine is dead, and can't be used for anything. And the amount of fuel in it was only a few hundred liters at maximum.