Solving Space Mechanics Homework: Dry Mass Fraction, F

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Homework Help Overview

The discussion revolves around calculating the dry mass fraction, F, in the context of space mechanics, specifically relating to the mass ratio required for a spacecraft to achieve a certain delta-v. Participants explore how variations in F affect the dry mass and payload capacity of a vehicle given a fixed amount of propellant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to understand the derivation of the payload mass based on the dry mass fraction and propellant mass. Questions arise about the calculations leading to specific values of payload and dry mass, particularly regarding the implications of different values of F.

Discussion Status

There is an ongoing exploration of the relationships between dry mass, propellant mass, and payload. Some participants provide mathematical expressions to clarify how these quantities interact, while others express confusion about specific values and their derivations. The discussion reflects a mix of interpretations and attempts to reconcile different understandings of the mass ratio and its implications.

Contextual Notes

Participants note that the mass ratio must be maintained at a specific value to achieve the desired delta-v, and there is uncertainty about how certain values are derived, particularly the consistent use of 10T for dry mass plus cargo when propellant is fixed at 90T.

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Homework Statement



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.

Homework Equations



\frac{M+P}{P}= e^{ΔV/M}

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

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:
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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?
 
Simon Bridge said:
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?

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:
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:
But if F = 0.11, the dry mass of the system would be 9.9 tonnes and the payload would fall ... to 0.1 tonne.

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.
 
Simon Bridge said:
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.



Does
\frac{90T + 9.9T + C}{9.9T + C} = 10T ?

Then solving for C
...

C = 0.1T
 
Biosyn said:
Does
\frac{90T + 9.9T + C}{9.9T + C} = 10T ?

Then solving for C
...

C = 0.1T
Now you know where that 0.1 comes from?
 
Simon Bridge said:
Now you know where that 0.1 comes from?

Yup, but uh..I'm not entirely sure where 10T comes from.

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