Coolant [Spaceships]

[malloyd]

Originally Posted by Johnny1A.2

The hypothetical Daedalus probe, for ex, if it struck a habitable world at its cruise speed of .12c, would be a monster. We're talking many hundreds of gigatons, at least. Probably more power than the combined arsenals of both superpowers at the height of the Cold War (about 60 gigatons IIRC).

Of course if you have a power plant that will get something up to planet cracking energies in less than a lifetime, it's not much of a threat to an equally capable target who have prepared for it. With that kind of energy to work with they can emit a radar pulse that will image it light-days out, which they use to target the beam weapon that will flash it to plasma in milliseconds at that same range.

You really do need to ignore the other things a civilization could do with vast energies to make it a center stage threat, probably starting with all the other more efficient ways we haven't thought of yet you could apply vast energies to devastating planets.

[malloyd]

Originally Posted by Ulzgoroth

Yes, applying the Tsiolkovsky rocket equation to the reactionless-but-delta-V-limited drive has the effective exhaust velocity you would use in the equation not actually corresponding to the velocity of anything in the system. But the equation still works exactly the same way.

Yes. There is a reason for the tradition of writing the equation and reporting performance in terms of specific impulse and not exhaust velocity. It's perfectly possible to apply slightly modified versions of the "rocket" equation to things like jet engines and helicopter rotors. You can get specific impulses that correspond to "exhaust velocities" greater than the speed of light when you do. Obviously nothing is actually moving anywhere near that fast, you are just using something other than the burned fuel to provide the momentum.

You don't need to consume a coolant to describe a reactionless thruster in terms of the rocket equation. If it consumes power you can compute a specific impulse in terms of consumption of whatever the power plant uses for fuel.

[Fred Brackin]

Originally Posted by Johnny1A.2

The hypothetical Daedalus probe, for ex, if it struck a habitable world at its cruise speed of .12c, would be a monster. We're talking many hundreds of gigatons, at least. Probably more power than the combined arsenals of both superpowers at the height of the Cold War (about 60 gigatons IIRC).

The number of strategic warheads I have was 30,000. The great majority of the US contribution was 1/3rd megaton warheads on Minutemen. The W84 I believe. The only US warheads larger than that would have been soem on the rather small Atlas and Titan fleets and on SAC bombers. This number was never very high compared to the Minutemen even before they went MIRV.

So I think your number is a considerable overestimate even allowingf for larger Soviet warheads.

The rating for any c fractional vehicle is easy to calculate. It's the square root of the velocity in %c x the mass in antimatter equivalent. At c it would be 100% antimatter equivalent and you've spent the equivalent of the vehicle's mass in a 100% conversion drive to get there. C is where the Einstein equation (e=mc2) and the KE equation (ke=mv2) intersect.

So 12%c gives you 3.46 x the mass in antimatter with every metric ton of antimatter equivalent equalling 43 gigatons of TNT.

A quick google on "Daedalus probe" gives a scientific payload of 500 tons so that's 1.73 teratons rather than gigatons.

[Agemegos]

Originally Posted by Fred Brackin

The rating for any c fractional vehicle is easy to calculate. It's the square root of the velocity in %c x the mass in antimatter equivalent. At c it would be 100% antimatter equivalent and you've spent the equivalent of the vehicle's mass in a 100% conversion drive to get there.

E = ½mv² is the non-relativistic approximation for kinetic energy, which is not good to use for objects colliding at a significant fraction of c. And even in that energy is proportional to the square of velocity, not the square root of velocity. The expression for kinetic energy in special relativity is

E = mc² {[1/√(1-v²/c²)]-1}

That increases without bound as v → c (goes to infinity at c, not to mc²).

[RyanW]

Nothing specific to add, except that Project Thor had one massive flaw:

Project Damocles is a much better name.

[Agemegos]

Originally Posted by malloyd

You don't need to consume a coolant to describe a reactionless thruster in terms of the rocket equation. If it consumes power you can compute a specific impulse in terms of consumption of whatever the power plant uses for fuel.

Provided that it dumps spent batteries or whatever rather than carrying them as a dead load, and has its performance improve as the load lightens.

[Anthony]

Originally Posted by Agemegos

Provided that it dumps spent batteries or whatever rather than carrying them as a dead load, and has its performance improve as the load lightens.

Yeah, the actual requirements for the rocket equation to work are that a specific quantity of lost mass produces a specific total impulse. The details of how the mass is lost are unimportant, but the rocket equation is just the solution of the integral dV = K*dM/M (where V is velocity and M is mass), so it doesn't work if K is not a constant or the thrust equation doesn't match the above.

[Agemegos]

Originally Posted by Anthony

Yeah, the actual requirements for the rocket equation to work are that a specific quantity of lost mass produces a specific total impulse. The details of how the mass is lost are unimportant, but the rocket equation is just the solution of the integral dV = K*dM/M (where V is velocity and M is mass), so it doesn't work if K is not a constant or the thrust equation doesn't match the above.

The rocket equation does not apply to an electric car travelling along a highway with rolling resistance, aerodynamic drag, and a speed limit.

[Anthony]

Originally Posted by Agemegos

The rocket equation does not apply to an electric car travelling along a highway with rolling resistance, aerodynamic drag, and a speed limit.

That's two separate issues. One is that the rocket equation only measures delta-V caused by the rocket, not from other sources, and the other is that vehicles that push on an external medium have variable performance based on speed relative to the medium they're pushing on.

[Agemegos]

Originally Posted by Anthony

That's two separate issues. One is that the rocket equation only measures delta-V caused by the rocket, not from other sources, and the other is that vehicles that push on an external medium have variable performance based on speed relative to the medium they're pushing on.

I count four issues. The battery doesn't get significantly lighter as it discharges. Speed doesn't change when force is applied to overcome resistance. Aerodynamic drag varies with speed. Electric motor and transmission efficiency vary with speed.

People sometimes assure me that Tsiolkovsky's equation applies to a hovering helicopter. Not if Δv = v(final) - v(initial) it doesn't.