Thank you. I knew I had an error somewhere, now I have an idea where to look, and something to check against. I appreciate your help greatly.

I will now refactor my sources.

]]>Fair criticisms both of you. Oops T = it’s seconds x seconds, sorry typo. It does cover deceleration though, probably because the ship needs to go twice as fast. The rest of Neutrino’s criticisms are spot on though.

Huh. I got my 8.8 from here.

]]>The distance is actually 8.6 light years, per here: http://en.wikipedia.org/wiki/Sirius

]]>Your units are inconsistent, KNO3. If D is in meters and T is seconds, A must be meters/second-squared. Units as well as numbers have to be equal on both sides of an equation. Your arithmetic yields 0.377 meters/sec2. Which matches mine (except that I used 8 LY even. I took http://ellieonplanetx.com/2010/06/21/comics/art-of-conversation/ as canon.)

Gamma of 1.1533 means that a kg (rest) traveling at that speed masses 1.1533 kgs. The extra 0.1533 kg is its kinetic energy. That much mass (M and AM half and half?) had to be converted to energy.

That’s an absolute minimum. A rocket requires MUCH more energy. The fuel has to be accelerated until it’s used. And most of the energy winds up in the exhaust jet. To have a sane mass-ratio you need high exhaust velocity — photons in the limit. But radiation carries very little momentum (which is what you want) per unit of energy.

And we haven’t even mentioned slowing down at the end. Ellie made a reasonably “hard” landing on Planet X — but not THAT hard. Again, details to anyone who asks.

]]>Dunno, getting different numbers.

Using the 30 year number is accurate to two significant figures.

Distance to Sirius is 8.8 light years, again two significant figures.

Constant acceleration

Accelerate to midpoint, flip over, then decelerate.

Formula: A=4D/T (dunno if this formula is right, it’s the only one I have)

A=G’s acceleration

D=Meters

T=Seconds

4 x 8.33E16 / 8.836E17 =

Yields .377 g’s acceleration times

9.80665 to get meters per second squared acceleration.

Which is 3.7 m/s squared (twice your figure)

Next part if from memory… so

3.7 m/s = 3.7 joules

3.7 joules to accelerate one Kg at 3.7 m/s squared

times 94 billion seconds (30 years)

Gives 3.5E11 joules

Given 1.8E14 Joules of antimatter per Kg

2 grams of antimatter? 86 Kiloton Nuke?

Huh.

I don’t think so.

If the average speed is 8/30 = 0.266 c then the top speed is double that, 0.533 c, because the speed is zero at both ends of the trip. This is simple Newtonian dynamics, of course.

Before I solve the Relativistic problem, I realize most of Ellie’s followers DON’T want to see the comments section devolve into a physics lecture. I’ll just summarize the results. If anyone’s interested – and The Institute agrees – I can e-mail Mission Control and let him forward a more complete analysis.

It’s easiest to break the problem into two halves. Crossing 4 lightyears in 15 years. That’s “rest time”, measured in an un-accelerated frame.

If was assume constant boost;

Peak gamma 1.15335503798433

Peak velocity 0.498246007471092 c

Proper acceleration 3.7130E-02 g or 0.3641236882 m/s2

Boost period (proper) 14.27 yrs

Boost period (rest) 14.9922611949376 yrs

Boost distance (rest) 4 lt-yrs

We can do MUCH better with non-constant boost. If the acceleration is high (but brief) at both ends, nearly the entire flight can be made coasting at 0.266 c. By cutting the total delta-vee in half, the required mass ratio goes WAY down. (How much depends on the exhaust velocity.)

]]>They were using constant acceleration?

]]>Can someone check my calculations? According to mine, the minimum energy required to get to Planet X in 30 years at constant acceleration (0.186177m/s²) is 15.1179 petaJoule per kilogram, double that if it is a reaction engine, not counting fuel. This would be equivalent to more than 168 grams of pure anti-matter per kg, not counting confinement. The engine would have to have a minimum power output of nearly 15.97 Megawatts (again double that if a rocket) per kilogram payload, not counting entropy and other losses.

The top speed would be circa 0.28214c (or 0.29c on-board, computed classically, relativistic effects stretching the distance for five months worth).

I expect the effect of the planet’s own velocity on these figures to be negligible.

How much am I off?

]]>