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jacobmuller 05-07-2010 03:24 AM

variable drive
 
I don't know if this is covered somewhere but I do know I don't know how to figure it out:)

A reactionless drive with thrust that varies with local gravity, eg low orbit 1G, 50,000 miles out 0.005G, per drive.

Problem for me being, there's formulae there for #G and distances etc but what happens to the math when G varies according to distance to/from a source?

I don't want a fixed reactionless drive system but if the math gets too painful I might settle for a "only works in significant G" fiat, ie only surface to orbit.

vierasmarius 05-07-2010 07:56 AM

Re: variable drive
 
Quote:

Originally Posted by jacobmuller (Post 978208)
I don't know if this is covered somewhere but I do know I don't know how to figure it out:)

A reactionless drive with thrust that varies with local gravity, eg low orbit 1G, 50,000 miles out 0.005G, per drive.

Problem for me being, there's formulae there for #G and distances etc but what happens to the math when G varies according to distance to/from a source?

I don't want a fixed reactionless drive system but if the math gets too painful I might settle for a "only works in significant G" fiat, ie only surface to orbit.

I'd suggest just setting the max thrust at some multiple of local gravity, probably x2 to x5 or so. That way you'd always be able to take off, but your engine power would drop quickly as you get farther away. I don't know the equation for what the gravity is further from the surface, but other folks here on the forum should be able to help with that.

lexington 05-07-2010 08:22 AM

Re: variable drive
 
If you know the mass of the planet and distance away from it's center of mass the acceleration toward the planet's surface is m/r^2 (where m is a multiple of Earth's mass and r the distance from center of mass divided by earth's mean radius). The answer is in gees.

So if the ship accelerates at 1G on the surface of Earth and very low orbit then at the edge of LEO (2000km) it would accelerate at 1/(1.3^2)=0.6G, assuming the the changes is proportional. At 50000 miles 1/(12.5^2)=0.006G.

If you're looking for travel times I believe you have to calculate jerk, which is a calculus problem.

jacobmuller 05-07-2010 09:09 AM

Re: variable drive
 
Yes, travel times, badly edited post, as usual...

Calculus - didn't do it, did do Physics*, hence problem. I'm tempted to study it but "yield not unto temptation" is burned into my subconscious.

If the opinion of those who have studied calculus is that doing this thing (drive G that varies as you travel to or from a mass) is just too painful for gaming I'll go with the "orbital booster" version and stick with plain old reaction drives for the deep beyond (high orbit+)

*"...Infinity. About fifty feet or so..."

malloyd 05-07-2010 09:40 AM

Re: variable drive
 
Quote:

Originally Posted by jacobmuller (Post 978299)
Yes, travel times, badly edited post, as usual...

Calculus - didn't do it, did do Physics*, hence problem. I'm tempted to study it but "yield not unto temptation" is burned into my subconscious.

I think the first year Calculus is going to be unavoidable on this one.
If the variation of acceleration with distance a(s) is a smooth function, you get a not too terrible equation for time to climb to height s.

dt^2 = (1/a(s)) d^2s

Integrate twice, plug in the desired distance and the radius of the planet as the limits and you're there. If you want a simple multiple of local gravity, then for a planet of radius r, t is going to be of the form t = k (s^4 - r^4), where k is a constant depending on the exact multiples you pick.


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