# Equations associated with calculating velocity of a bullet.

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1. Feb 28, 2016

### Teslanumber1

Hello, I'm currently doing a project on a Verne gun, and I'd just like to know which equations I should know in order to explain such a device. How do i calculate velocity, efficiency of the launch, how the gas laws play into the efficiency, since it should be traveling at 10s of kilometers a second what effect does that have on the air in front of the payload, since nuclear weapons will be used how much of the payload can I expect to vaporize, those sorts of things. I've been researching for a while now, and can't seem to find anything.

2. Feb 28, 2016

### Simon Bridge

That's because it's very complicated... too many variables.
eg. the amount of payload to vaporize depends on it's shape and what the outer casing is made out of.
So you need to know about atmospheric heating, ablation, and material properties. Probably a projectile that survives the launch would survive anything the atmosphere can do to it.

Your expected speeds are above escape velocity (about 11.2kmps), so you need to know about orbital dynamics.

For efficiency of launch, first you have to define it.
Energy efficiency would be the kinetic energy of the emerging projectile divided by the total energy that went into the shot.
The gun is working as a heat engine - but you have not said anything about the details of operation so we don't know what kind. So you'll need to know about those too... but that is how the gas laws factor in. BTW: the regular gas laws you may know from school likely won't be very important. You'll want to know a bit of plasma physics.

Mostly you need a detailed description of the process you want to use, and what outcome you are after.

3. Feb 28, 2016

### Teslanumber1

let the shape be a cylinder 24m in diameter 52 meters tall with a 4 meter inverted cone at the bottom and a 4 meter cone at the top. also the launch tube is 6km long with 1km just for water as a medium to decrease the amount of melting/vaporization of the payload, and the other 5km is a complete vacuum. the water would heat up 82,000 kelvin(so plasma, so yes gas laws wouldnt help), assuming perfect transfer of energy. the warhead used would be a 25 megaton bomb, the three stage b41 bomb to be specific. lets also say the average densisty of the payload is 13,000kg/m^3. also assuming perfect transfer of energy the velocity should be around 23.5km/s. lets also say the bottom 6 meters is composed of entirely tungsten. when i say efficiency i mean how much of the total energy of the bomb is aborbed by the payload, then from there we get velocity, and then how much of that velocity does it retain at 400km up. also the payload is 272,155,422 kg. lets also ignore the payload shattering, or cracking from the stress of the explosion. Edit--this might be a dumb question, but would there be anyway to rifle this so it would be more stable, or would the rifling vaporize, or turn to dust due to the high speed of the payload?

Last edited by a moderator: May 7, 2017
4. Feb 29, 2016

### Simon Bridge

This is a very big vehicle just to deliver a nuke.

... seals and stuff to keep the vacuum,
Have a look at what it takes to build a big vacuum chamber:
https://en.wikipedia.org/wiki/Space_Power_Facility

... how did you work this out? Oh I see you have a density for the payload, so a payload mass is available and you know how fast you want it to go?
Note: the surface of the Sun is around 6000K.

... from what to where? You want energy to go into the vehicle ... so do we picture an explosive sitting between the vehicle and the water? Under the water?

Is this the bomb to be delivered?

This thing will basically just keep going. Note: escape velocity from the Solar System is 42.1km/s.

What are you thinking of making this it out of? The density of steel is about 8000kg/m^3, Osmium is 22000kg/m^3.
I make your vehicle around 300,000T of really expensive, high density, material.

Considering the usual cost to deliver a B41 bomb, is this a cost effective way to blow something up?

The vehicle is going to be riding the pressure wave rather than just absorbing energy and shooting off.

The blast needs around 90GJ ... a bit more than 2T of TNT ... to get converted to kinetic energy in the vehicle.
This is being detonated in a tube of some kind so it is hard to figure out how the shockwave will go ... some will be absorbed by the walls, some by water.
It's altered for shaped charges etc.

Air resistance will be a big factor ... but only for a short time: the atmosphere is less than 100km thick. Even if half the speed is lost passing through it, you have still exceeded escape velocity. This is not a surface-to-surface weapon. Thinking of targeting the Moon?

also the payload is 272,155,422 kg.[/quote]
You may want to check that.
Code (Text):
> 13000*pi*52*12^2
ans =    3.0582e+08
I doubt it would make much difference... the gun is basically one shot.

5. Feb 29, 2016

### Teslanumber1

You may want to check that.
Code (Text):
> 13000*pi*52*12^2
ans =    3.0582e+08

I doubt it would make much difference... the gun is basically one shot.[/QUOTE]
just a clarification here, the nuke is the propulsion method.

Last edited by a moderator: May 7, 2017