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Homework Help: Space Mechanics

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data

    The dry mass fraction, F when multiplied by the propellant mass gives the dry mass of the vehicle (not counting its payload). This dry mass faction F, is a function of system design and lightness of the materials employed for its construction. So if F = 0.1, a ship carrying 90 tonnes of propellant would have a dry mass of 9 tonnes. If the mass ratio of the system had to be 10 to perform a certain ΔV, this would allow the craft to carry 1 tonne of payload. But if F = 0.11, the dry mass of the system would be 9.9 tonnes and the payload would fall by a factor of 10, to 0.1 tonne. If F = 0.12, the dry mass of the vehicle would be 10.8 tonnes, and the mission would be impossible even without any payload.

    I need help understanding where 0.1 tonne comes from. I know how they got 1 tonne of payload.

    2. Relevant equations

    [itex]\frac{M+P}{P}[/itex]= e[itex]^{ΔV/M}[/itex]

    M = Dry mass of vehicle without payload
    P = Mass of propellant

    3. The attempt at a solution

    Let x = payload

    (9 + 90)/9 - X = 10

    X = 1 tonne

    (9.9 + 90)/9.9 - X = 10

    X = 0.09090909~ tonne
    Last edited: Oct 22, 2012
  2. jcsd
  3. Oct 22, 2012 #2

    Simon Bridge

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    You mean the 0.1T cargo for F=0.01?

    mass ratio has to be 10 to make a delta-vee.
    then a ship with 90T of fuel can still do the delta-vee if the combined dry-mass and cargo-mass is smaller than a certain amount. What is this amount?

    What is the dry-mass for a ship with F=0.01 and 90T of fuel?
    How much does this leave for cargo?
  4. Oct 22, 2012 #3
    Sorry, where did you get F=0.01 from? I'm confused.

    The amount of payload for a ship with F=0.1 and 90T of fuel will be 1T.

    I'm not sure where they came up with 0.1T of cargo for a ship with F=0.11 , 90T of fuel , and 9.9T of dry mass.

    The mass ratio is 10 = (M+P)/M right?
    Last edited: Oct 22, 2012
  5. Oct 22, 2012 #4

    Simon Bridge

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    F=0.11 is the example given but you seem to have found that confusing. The math is the same for any F. Here's the bit I guessed we were talking about:
    mass ratio = R
    mass of fuel = P
    dry mass = D
    cargo mass = C
    then mass-ratio is R = (P+D+C)/(D+C)

    so for P=90T, D+C=10T to get a mass-ratio: 100/10 = 10
    as a consequence: C = 10-D.

    If you like: since F=P/D, R=[(1+F)P+C]/[FP+C] solve for C
    But the above example does it in 3 steps ... thus:

    1. Given R and P you get D+C.
    2. Given F and P you get D
    3. combine 1 and 2 to get C.
  6. Oct 22, 2012 #5

    [itex]\frac{90T + 9.9T + C}{9.9T + C}[/itex] = 10T ?

    Then solving for C

    C = 0.1T
  7. Oct 22, 2012 #6

    Simon Bridge

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    Now you know where that 0.1 comes from?
  8. Oct 23, 2012 #7
    Yup, but uh..I'm not entirely sure where 10T comes from.

    Is D+C always going to be 10T when P=90T?

    Or is it C = 0.1 because the units cancel out. Therefore that's why the Ratio is equal to [(1+F)P+C]/[FP+C].
    Sorry, I think I'm just going around in circles. ._.
    Last edited: Oct 23, 2012
  9. Oct 25, 2012 #8

    Simon Bridge

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    There's a bunch of definitions:
    Dry mass, and mass ratio. These are separate.
    The 10T comes from a mass-ratio of 10, when the propellant mass is 90T.
    (You'd think a MR of 10 would mean "10:1" wouldn't you? But it doesn't.)

    Using above notation:
    if M = D+C+P is the total mass of the craft.
    MR = M/(M-P)

    So a 100T ship with MR=10 has P=90T and D+C=10T.

    Rocket design is actually quite complicated and these wee numbers help engineers make design decisions.
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