Rail gun damage?

[gewing]

Well, using megaplayboy's 1/2000th of c, and an average density of 7 x 10^-24 g/cc (which actually includes suns, planets, etc.---sorry, I didn't write down the source for that number), you will have a momentum of 1.573 x 10^-14 kg-m/s for every square centimeter of cross-sectional area of the projectile, every second. You'll have to determine the mass and cross-section of the projectile; the first to determine the projectile's forward momentum, the second to find the "pushing back" momentum.

Of course, the figure of "pushing back" momentum will drop with velocity, as you'll be going through less space, and running into less mass each second.

To pull an example out of thin air: A cylinder of lead one meter in diameter and three meters long masses about 26,743 kg. At 1/2000th c its momentum is ~4.009 x 10^9 kg-m/s. The cross-sectional area is about 7854 sq cm, so the "pushing back" momentum is 1.235 x 10^-10 kg-m/s. This is about 3.082 x 10^-20 of the projectile's momentum. IOW, if the resisting momentum and "forward" momentum stayed constant, it would take ~ 3.25 x 10^19 seconds (1.02 trillion years) to bring the projectile to a stop.

Or, in short, "It ain't never gonna stop."

that is an average,and very interesting, but I suspect not very practical, really.

Though I will not even ATTEMPT to find the mathematical models I suspect that it would be subject to diversion due to gravitational forces even if it misses the star or planet, other diversions due to solar wind effects, and if it hits one of the more solid bits of the universe (even a dust speck, weighing one grain(1/7000 of a pound) would probably vaporize a significant portion of the mass of said cylinder) things would get very messy very fast!!! I get
2145989401510 ft lbs. I rather hope I missed something.

[Cancer]

Renegade Legion used Spinal Mounts, as does HERO: Combat Evolved. Both the In Amber Clad and Pillar of Autumn (a light Frigate and a medium cruiser, respectively) use a 16d6 Spinal Mount with limited ammunition and likely a ridiculous range. One wonders, if fired in space where there's no gravity, how long it would take the thing to slow down to the point where it won't annihilate something else in its path?

To supply the physics:

the pressure on the projectile will be 0.5 * (density of medium) * (relative velocity)^2

So the total force on the projectile is (pressure) * (cross-sectional area of projectile)

And the instantaneous deceleration of the projectile is that force divided by the projectile mass. You have integrate over time to get the effect on the velocity (or just multiply if you assume a constant medium density).

The density of gas in space varies, of course, but you can take 1 atom per cubic centimeter as a representative number. The mean molecular weight in space is about 1.4 AMU. You can assume the gas is at rest so the relative velocity is the projectile velocity. The cross-sectional area of the projectile is just the area of the projectile as you see it come at you .

Now plug in numbers as you want.

Thing is, just about all collisions between objects in space are going to be hypersonic (that is, the relative speed of the colliding objects will be greater than the speed of sound within either object). The random distribution of velocities in space is on the order of several km/sec, which is amply destructive by itself.

Basil's point is right: it'll take a long time for gas ram pressure alone to affect the velocity. Electromagnetic effects will likely be unimportant as well (since the conducting projectile is passing through magnetic fields, there'll be eddy currents working to oppose the motion ... which'll be trivial in magnitude). But, over that time, the gravitational effects of stars, giant molecular clouds, black holes, the collective galaxy, the dark matter, etc., will alter the projectile velocity (both speed and direction).

At some point with close-range combat the real function of the railgun is to throw lethal slugs at a high enough speed that the opponent can't react fast enough, and get out of the way or destroy the incoming projectiles quickly enough to avoid destruction.

[megaplayboy]

at 1/2000 of light speed, it would take 12 days just to traverse 1 AU of distance, and of course 2000 years to cover a light year. Not a very effective means of long distance bombardment--firing out from the fringe of the solar system(about 40-50 AU), it'd take over a year for a projectile to hit Earth. But from the moon (about 450,000 km), it could make the trip in under an hour.

[Thia Halmades]

To supply the physics:

the pressure on the projectile will be 0.5 * (density of medium) * (relative velocity)^2

So the total force on the projectile is (pressure) * (cross-sectional area of projectile)

And the instantaneous deceleration of the projectile is that force divided by the projectile mass. You have integrate over time to get the effect on the velocity (or just multiply if you assume a constant medium density).

The density of gas in space varies, of course, but you can take 1 atom per cubic centimeter as a representative number. The mean molecular weight in space is about 1.4 AMU. You can assume the gas is at rest so the relative velocity is the projectile velocity. The cross-sectional area of the projectile is just the area of the projectile as you see it come at you .

Now plug in numbers as you want.

Thing is, just about all collisions between objects in space are going to be hypersonic (that is, the relative speed of the colliding objects will be greater than the speed of sound within either object). The random distribution of velocities in space is on the order of several km/sec, which is amply destructive by itself.

Basil's point is right: it'll take a long time for gas ram pressure alone to affect the velocity. Electromagnetic effects will likely be unimportant as well (since the conducting projectile is passing through magnetic fields, there'll be eddy currents working to oppose the motion ... which'll be trivial in magnitude). But, over that time, the gravitational effects of stars, giant molecular clouds, black holes, the collective galaxy, the dark matter, etc., will alter the projectile velocity (both speed and direction).

At some point with close-range combat the real function of the railgun is to throw lethal slugs at a high enough speed that the opponent can't react fast enough, and get out of the way or destroy the incoming projectiles quickly enough to avoid destruction.

First, wow, thank you (and Basil) both for having the math I lack. Writing a book? I'm your man! Doing a math problem! No sweat! Higher than Algebra? Er... wait...

And more importantly, thank you (again, both, I believe I owe one of you rep at this point, I think I got Basil already) for putting it in plain English that my mind can wrap around. And yes, I agree that the purpose of a rail weapon/coil gun on a cap ship is to put holes in other cap ships at SHORT ranges, not long ones, although it can do that, it isn't the primary intent of the design. Again, awesome, thank you both.

[megaplayboy]

For reference purposes, I think the highest g-forces projected for gauss weaponry are on the order of 250,000 gees.

Since we can posit a length d for any given weapon "caliber"(and I'm assuming that number will be between 100 and 1000x the slug bore--IOW, a 20mm gauss weapon will have a barrel length between 2m and 20m), and we have an upper limit for a(about 2.5 x 10E6), then we can solve for t in the equation
D(distance travelled) = 1/2 x a(acceleration) x t^2(time).

If we apply a as a constant 2.5 x 10E6 to the equation, then
D = 1.25 x 10E6 x t^2. A little more work, and we get t^2 = D/1.25 x 10E6

For a given length and acceleration, we can calculate the velocity generated, since v=at.
For D = 125 m, t^2 = 1/10E4, which means t= 1/100 sec. Multiply through, and you get v = 2.5 x 10E6 x 10E-2, or 2.5 x 10E4 meters/sec. (25km per second, a very respectably fast projectile speed for a cruiser-sized spinal mount gauss weapon).

[Nyrath]

Cancer can correct me if I am wrong, but the basic equation is

If the relative velocity between the railgun/coilgun round and the target is below 14% of the speed of light:
K = 0.5 * M * V^2
where
K = kinetic energy in joules
M = mass of round in kilograms
V = relative velocity in meters per second
x^2 = the square of x (i.e., x times x)

(non-physics people can replace the term relative velocity with "muzzle velocity" for most situations)

At first approximation, you can assume that all the kinetic energy creates damage to the target (for second approximation, you have to consider the possibility that the round drills a hole in the target and exits the far side with wasted damage). Divide joules by 4.184 x 10^6 (i.e., 4184,000) to get the equivalent damage in kilograms of TNT. Divide by 4.184 x 10^12 for kilotons, and 4.184 x 10^15 for megatons. Or you can refer to the Boom Table.

If the projectiles are traveling faster than about 14% lightspeed, then you have to start taking Einstein's Relativity into account (i.e., they are now "relativistic weapons"). The equation is:
K = ((1/sqrt(1 - (V^2/9e16))) - 1) * M * 9e16
or
K = ((1/sqrt(1 - P^2)) - 1) * M * 9e16
where
K = relativistic kinetic energy (Joules)
M = mass of projectile (kg)
V = velocity of projectile relative to target (m/s)
P = velocity of projectile relative to target (percentage of c, e.g., three quarters lightspeed = 0.75)
sqrt(x) = square root of x
9e16 = 90,000,000,000,000,000

Examples:

Nightlord256's MAC gun fires a 3,000 tonne (3,000,000 kg) projectile at 0.35 c.
K = ((1/sqrt(1 - P^2)) - 1) * M * 9e16
K = ((1/sqrt(1 - 0.35^2)) - 1) * 3e6 * 9e16
K = ((1/sqrt(0.8775)) - 1) * 2.7e23
K = 0.0675 * 2.7e23
K = 1.823e22 joules
K = 4.4 teratons
K = 101 metric tons of antimatter
K = 1657 earthquakes measuring 9.5 on the Richter scale
K = 28,929 Krakatoas

megaplayboy's 150 km/sec projectile might have energies in the megaton range if the projectile mass is high enough. How much mass for 1 megaton?
K = 0.5 * M * V^2
M = K / (0.5 * V^2)
M = 4.184e15 / (0.5 * 1.5e5^2)
M = 371,900 kg = 372 metric tons

[Thia Halmades]

*blinks rapidly and grins like an idiot*

... what?

[Nyrath]

*blinks rapidly and grins like an idiot*

... what?

Well, in simple terms:

1. Start with the Boom Table and figure how how many joules corresponds to how many Hero dice of damage (use some kind of logarithmic scale)
2. Use either of the two equations I posted to calculate how many joules of damage a projectile weapon inflicts
3. Look up that amount of joules on the table you made in step [1] to see how many dice of damage it is.

[Kristopher]

Ugh.

I don't know how I feel about doing direct energy-to-damage comparisons. That's a bit too close to the conversions people tried to do to for Classic Battletech, which were admittedly fatally flawed because in that game, a large mech punches for as much game-defined damage as is done by a hit from a an anti-mech weapon.

[megaplayboy]

Cancer can correct me if I am wrong, but the basic equation is

If the relative velocity between the railgun/coilgun round and the target is below 14% of the speed of light:
K = 0.5 * M * V^2
where
K = kinetic energy in joules
M = mass of round in kilograms
V = relative velocity in meters per second
x^2 = the square of x (i.e., x times x)

(non-physics people can replace the term relative velocity with "muzzle velocity" for most situations)

At first approximation, you can assume that all the kinetic energy creates damage to the target (for second approximation, you have to consider the possibility that the round drills a hole in the target and exits the far side with wasted damage). Divide joules by 4.184 x 10^6 (i.e., 4184,000) to get the equivalent damage in kilograms of TNT. Divide by 4.184 x 10^12 for kilotons, and 4.184 x 10^15 for megatons. Or you can refer to the Boom Table.

If the projectiles are traveling faster than about 14% lightspeed, then you have to start taking Einstein's Relativity into account (i.e., they are now "relativistic weapons"). The equation is:
K = ((1/sqrt(1 - (V^2/9e16))) - 1) * M * 9e16
or
K = ((1/sqrt(1 - P^2)) - 1) * M * 9e16
where
K = relativistic kinetic energy (Joules)
M = mass of projectile (kg)
V = velocity of projectile relative to target (m/s)
P = velocity of projectile relative to target (percentage of c, e.g., three quarters lightspeed = 0.75)
sqrt(x) = square root of x
9e16 = 90,000,000,000,000,000

Examples:

Nightlord256's MAC gun fires a 3,000 tonne (3,000,000 kg) projectile at 0.35 c.
K = ((1/sqrt(1 - P^2)) - 1) * M * 9e16
K = ((1/sqrt(1 - 0.35^2)) - 1) * 3e6 * 9e16
K = ((1/sqrt(0.8775)) - 1) * 2.7e23
K = 0.0675 * 2.7e23
K = 1.823e22 joules
K = 4.4 teratons
K = 101 metric tons of antimatter
K = 1657 earthquakes measuring 9.5 on the Richter scale
K = 28,929 Krakatoas

megaplayboy's 150 km/sec projectile might have energies in the megaton range if the projectile mass is high enough. How much mass for 1 megaton?
K = 0.5 * M * V^2
M = K / (0.5 * V^2)
M = 4.184e15 / (0.5 * 1.5e5^2)
M = 371,900 kg = 372 metric tons

with a high density projectile, that'd be about 20 cubic meters of projectile--maybe a gauss weapon with a 1 meter diameter bore and a 20 meter long dense rod?

Up the diameter and length by a factor of 10(you'd need a ludicrously long "barrel"), and you've got gigaton level "rods from God"--I'm pretty sure the surface damage from one of those would be highly "consequential".

[Thia Halmades]

Well, in simple terms:

1. Start with the Boom Table and figure how how many joules corresponds to how many Hero dice of damage (use some kind of logarithmic scale)
2. Use either of the two equations I posted to calculate how many joules of damage a projectile weapon inflicts
3. Look up that amount of joules on the table you made in step [1] to see how many dice of damage it is.

YAY! Simple terms! Thanks Nyrath!

[megaplayboy]

So, after looking at the boom table page, I have a tangentially related question: what would the tensile strength, radiation resistance and vaporization threshold(is is joules per cm^2?) of a hypothetical supermaterial have to be to withstand a 1 kiloton fission bomb blast at a range of 1 meter? Any way this can be calculated?--I saw a nuclear weapon effects calculator on there somewhere...

I mean "withstand" in the same sense one would use it for evaluating the survivability of an object and it's protective value to those inside it.

[Nyrath]

with a high density projectile, that'd be about 20 cubic meters of projectile--maybe a gauss weapon with a 1 meter diameter bore and a 20 meter long dense rod?

Up the diameter and length by a factor of 10(you'd need a ludicrously long "barrel"), and you've got gigaton level "rods from God"--I'm pretty sure the surface damage from one of those would be highly "consequential".

That's putting it mildly. One gigaton would make Krakatoa look like a mild burp.

Asteroidal material is abundant and readily available, though you'd have to refine it a bit to up the density.

[Nyrath]

So, after looking at the boom table page, I have a tangentially related question: what would the tensile strength, radiation resistance and vaporization threshold(is is joules per cm^2?) of a hypothetical supermaterial have to be to withstand a 1 kiloton fission bomb blast at a range of 1 meter? Any way this can be calculated?--I saw a nuclear weapon effects calculator on there somewhere...

Was it this one?
http://www.5596.org/cgi-bin/nuke.php
This one is not as useful
http://www.stardestroyer.net/Empire/Science/Nuke.html

Figuring armor is tricky. If the bomb detonates here, on Planet Earth, in the atmosphere, the explosion produces blast (overpressure), thermal (heat), and nuclear radiation
http://www.atomicarchive.com/Effects/effects1.shtml
If the bomb explodes in deeps space, all the energy is in radiation.

[tkdguy]

Useful stuff here, guys. Thanks.

I think I'll look at ram accelerators as well.