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Old 10-03-2017, 03:19 AM   #1
jacobmuller
 
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Default [Spaceships] simply fuel math...

RAW Super Fusion Torch, 50G 450mps but, if you allow use of an SM-4 section to get only 0.5G, how far will that tank of fuel take you - 45,000mps?

This is either one of those math things so simple you can't believe your own answer or it's Costello's rent arrears and I'm the landlord:/
I mean, I can't believe my own answer so, I need someone else to either say "Yup, it's really that simple." or swat me upside the head and show me the actual math that proves it's really all rocket science.

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Old 10-03-2017, 03:23 AM   #2
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Default Re: [Spaceships] simply fuel math...

Quote:
Originally Posted by jacobmuller View Post
RAW Super Fusion Torch, 50G 450mps but, if you allow use of an SM-4 section to get only 0.5G, how far will that tank of fuel take you - 45,000mps?

This is either one of those math things so simple you can't believe your own answer or it's Costello's rent arrears and I'm the landlord:/
It's simple the other way around. It gets you 450 mps. No matter what size the engine is, the final velocity depends on the size of the fuel tank relative to the entire ship, not relative to the engine. A smaller engine just takes longer to get there - 40 hours instead of 24 minutes to burn your tank dry.
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Old 10-03-2017, 04:31 AM   #3
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Default Re: [Spaceships] simply fuel math...

Looked at from a different angle, adding more engines doesn't get you any less delta-V from the single fuel tank, so removing fractions of an engine shouldn't get you any more.
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Old 10-03-2017, 05:00 AM   #4
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Default Re: [Spaceships] simply fuel math...

Delta-V - the change in velocity - is a measure of engine efficiency and does not change with engine size (at least, not on the scale of engines mounted on spaceships). A larger engine uses fuel just as efficiently as a smaller engine does.
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Old 10-03-2017, 05:35 AM   #5
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Default Re: [Spaceships] simply fuel math...

Thank you, all.
Brain off day. I think part of my scribbled notes were what if you "super" the rocket and scale from there and then I've wandered off into Lalaland:D
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Old 10-03-2017, 07:47 PM   #6
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Default Re: [Spaceships] simply fuel math...

Ok, piggy backing on this thread...

Suppose you decide to design a spaceship with the following in mind:

You have a SM 9 ship. A single "reaction Engine" for this ship will produce 3G acceleration.

However, using the Swapping out rule from page 5 of Transhuman Space ships...

Suppose I decide to use one SM 8 rocket engine to produce 1G extra, allowing for a 4G acceleration?

Further suppose that I use the other two modules for Fuel. Those two fuel will be 2/3rds of a fuel tank for a SM 9 ship.

So what happens to the Delta-V increase (page 17 of the First GURPS SPACEBOOKS rules)?

It states that if you have 6 to 8 tanks, multiply your delta-V by 1.2. If you have 9 to 12, multiply by 1.4

Now, if we have 2/3rds of a tank, we have 2/3rds of the delta-V that a normal SM9 tank would have. No question there.

But what happens to the Delta-V multiplier when you're at the "threshold" level between the multiplier values? In other words...

if you have 5 and 2/3rds of a tank, does this get the full 1.2 modifier? If you have 8 and 2/3rds of a tank, do you get the full 1.4 multiplier?

How would you resolve that issue?
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Old 10-03-2017, 08:16 PM   #7
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Default Re: [Spaceships] simply fuel math...

While the initial question is well answered, I'd add a thought - the smaller engine will take 100 times as long to burn through the reaction mass...but since it's producing 1/100 times as much acceleration, naturally the delta-V comes out the same.
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Originally Posted by hal View Post
Ok, piggy backing on this thread...

Suppose you decide to design a spaceship with the following in mind:

You have a SM 9 ship. A single "reaction Engine" for this ship will produce 3G acceleration.

However, using the Swapping out rule from page 5 of Transhuman Space ships...

Suppose I decide to use one SM 8 rocket engine to produce 1G extra, allowing for a 4G acceleration?

Further suppose that I use the other two modules for Fuel. Those two fuel will be 2/3rds of a fuel tank for a SM 9 ship.

So what happens to the Delta-V increase (page 17 of the First GURPS SPACEBOOKS rules)?

It states that if you have 6 to 8 tanks, multiply your delta-V by 1.2. If you have 9 to 12, multiply by 1.4

Now, if we have 2/3rds of a tank, we have 2/3rds of the delta-V that a normal SM9 tank would have. No question there.

But what happens to the Delta-V multiplier when you're at the "threshold" level between the multiplier values? In other words...

if you have 5 and 2/3rds of a tank, does this get the full 1.2 modifier? If you have 8 and 2/3rds of a tank, do you get the full 1.4 multiplier?

How would you resolve that issue?
So (A) there's no actual reason to involve the smaller rocket in this, it has no bearing on the question.

(B) Assuming I was using the by-the-book delta-V rule rather than using the rocket equation to smooth it, I'd use it by the book - 5 and 2/3s is not 6 to 8, it is less than 6, so you don't get that multiplier. 8 and 2/3s is slightly more of a problem, but for consistency I'd put it with the lower group rather than the upper group.
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Last edited by Ulzgoroth; 10-03-2017 at 08:21 PM.
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Old 10-03-2017, 09:54 PM   #8
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Default Re: [Spaceships] simply fuel math...

Quote:
Originally Posted by hal View Post
if you have 5 and 2/3rds of a tank, does this get the full 1.2 modifier? If you have 8 and 2/3rds of a tank, do you get the full 1.4 multiplier?
OK, if you want something more accurate, but more fiddly, then I think this is right. Take the natural logarithm of 20 divided by 20 minus the number of fuel tanks and divide by 0.05 (ln(20/(20-tanks))/0.05). This gives you the effective number of fuel tanks for that particular drive, so calculate your delta V from there.

Aside: The multipliers given in the book exist to approximate the rocket equation's effect on Delta V (i.e. higher proportion of propellant, higher delta V). However, they do seem to error on the low (but occasionally the high). For example, 19 tanks would have a multiplier of 3.15 according to my equations, but it gets a multiplier of three. On the other hand, 6 tanks should have a multiplier of 1.18 according to my equations, but it gets a multiplier of 1.2.
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Old 10-03-2017, 10:27 PM   #9
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Default Re: [Spaceships] simply fuel math...

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Originally Posted by TGLS View Post
OK, if you want something more accurate, but more fiddly, then I think this is right. Take the natural logarithm of 20 divided by 20 minus the number of fuel tanks and divide by 0.05 (ln(20/(20-tanks))/0.05). This gives you the effective number of fuel tanks for that particular drive, so calculate your delta V from there.
It's probably simpler to describe "divide by 0.05" as "multiply by 20".
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Old 10-03-2017, 10:43 PM   #10
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Default Re: [Spaceships] simply fuel math...

Quote:
Originally Posted by TGLS View Post
OK, if you want something more accurate, but more fiddly, then I think this is right. Take the natural logarithm of 20 divided by 20 minus the number of fuel tanks and divide by 0.05 (ln(20/(20-tanks))/0.05). This gives you the effective number of fuel tanks for that particular drive, so calculate your delta V from there.
If you want to be really pedantic, divide by ln(20/19) instead of 0.05. Assuming the table is accurate for exactly one tank, that will give you 1 = 1. It's only about a 2.5% difference, though.
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