Calculating Length of Barrel/Rails for Basic Newtonian Space Physics

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In summary, Cyberduke was trying to explain why it is difficult to understand a concept. He said that it is difficult to apply the concept to a real-world situation. He used an example of hitting something with a hammer in deep space with little gravity. The hammer would be slowed down by the little gravity, and the impact would not be as powerful. He said that the same principle applies to a rail gun. The rail gun will have a maximum velocity, and once the projectile falls below the mass it cannot push the rail gun any further.
  • #1
Cyberduke
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Basic Newtonian space physics

Ok I am just making a simple space war simulation game for kicks, using C#

I wanted to build in some customization and let the player pick different guns (think rail guns)

picking the:
-length of the rails (m)
-average force applied to the projectile along the rails (N)
-mass of the projectile (kg)

and the mass of the ship in (kg) will be allready known

ok so F = MA
A = F/M ?

I have tried modeling the acceleration of the projectile and stoping it when the velocity would have ment the projectile position exceeded the length of rails choosen.

but is there a formula I can use to calculate final "muzzle velocity"?
and how can I tell how far the ship would move in the opposite direction?
Is there a better way I could do this?

can someone point me in the right direction please.


Cheers,

Cyberduke
 
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  • #2
I am just going to add to this something else I can't work out.

Been looking around on the web and these forums for most of today so far...

Inertia... I was just working out kinetic energy of projectiles and due to the velocity being of greater importance then mass.

KE = (0.5 * M) * (V * V)

The smaller the projectile propelled with the same force the more kinetic energy it has. Fine, but when I thought about it that didn’t feel right if you talk about firing grains of sand and punching a tiny hole though foot thick steel plate. (I suppose that can happen but...)

So inertia... I am guessing that I have to take the mass of the target and the mass of the projectile into account and subtract energy from the conversion (collision) somewhere along the line...

Please bear with me... I know these are stupid questions to some here but the only physics I have ever been taught was in bottom stream science class in school 10 years ago. (Dyslexic)
 
  • #3
Ok a test...
I had a section of oak scrap wood about half a meter long 10 cm wide and 4 cm thick

I proped it up on some bricks in my garden and took out 3 hammers

First I started hitting the wood in the middle with a http://www.steptoesantiques.co.uk/acatalog/vintage_toffee_hammer.1156.JPG"
I hit the wood 30 times. This did not look like it was doing anything at all not even denting the surface of the wood.

Then I picked up a http://www.refocus-now.com/Stock/CATII/images/0%20A%20Hammer%20and%20nails,%20banging%20a%20nail%20into%20a%20plank%20of%20wood%20s..jpg.jpg"
I hit the wood 15 times. This made lots of circular dent marks in the surface of the wood (but still a bit frustrating)

Finaly I picked up my http://www.tradecounterdirect.com/prodimages/sladge-hammer.jpg"
One good wack and I got a satisfiying cracking noise
second wack and it split the wood in half(ish) kinda split up the length.


My point is I could have hit the wood all day with the toffee hammer and it would have still done nothing.
In which case the sledge hammer would have done far more dammage for less energy used.

What principle I am I missing In the woods ablilty to absorb energy/damage?
would the same be true on a larger scale if instead of wood it was steel plate?


Thanks,

Cyberduke
 
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  • #4
#3 -- What about about the weights of the hammers? How far did you lift each one? How fast did they come down at the point of the impact? Does this help?
 
  • #5
cyberduke, hitting something with a hammer is a bad judge. For the most part, you aren't accelerating the light hammers to their fullest; instead, they're capped by your arms' ability to move. In addition, I bet you used two hands for the sledgehammer, and one for the light ones, and probably gave it a larger swing. In addition, gravity itself accelerates everything at the same rate, so you weren't working under the same conditions.

You'll encounter a similar problem (well, several problems) if you try using a grain of sand on a rail gun. First, can it actually FIT in the rail gun? If the bars are so close together to hold a grain of sand together, can they push with the same force as a pair of rails holding up a bullet? Second, is air resistance. The sand will fly out and sloooooooooowwwwww down. In addition (I may be wrong on this, but it seems right), most rail guns cap off at a maximum velocity, because the pushing up the bars only goes so fast. Once you get below the mass for this, it doesn't really help to make the projectile lighter
 
  • #6
ok maybe that was a bad example, since there are lots of other factors.

I was trying to explain why having trouble to get to grips with a concept
that I can actualy go and do.

Ok Imagen in deep space (very little gravity) there are two spheres of iron both with a mass of say 1,000 kg

there are then two iron projectiles one has a mass of 0.1 kg and the other 100kg

the 0.1 kg projectile is propelled at one of the 1,000 kg spheres useing 100,000 N of force for 1 second
A = F/M ?
sooo. 0.1 kg projectile travelles towards the 1000 kg sphere at 1,000 km/s

and the 100 kg projectile is propelled at the other sphere useing 100,000 N of force for 1 second sooo... at 1 km/s

the 0.1 kg projectile has (0.5 * 0.1) * (1000000 * 1000000)
= 50,000,000,000 KE

but the 100 kg object only has (0.5 * 100) * (1000 * 1000)
= 50,000,000 KE

Will both projectiles move their targeted spheres the by the same amount / do the same dammage apon impact?

I think I am missing something in these calculations and that is Inertia.
but I don't know how to factor it in.

missing for both for propelling the projectiles and for the impact.
 

FAQ: Calculating Length of Barrel/Rails for Basic Newtonian Space Physics

How do I calculate the length of a barrel or rails for basic Newtonian space physics?

To calculate the length of a barrel or rails for basic Newtonian space physics, you will need to consider several factors such as the desired velocity of the projectile, the force applied by the propellant, and the mass of the projectile. Using Newton's second law of motion (F=ma), you can determine the necessary length to achieve the desired velocity.

What is the formula for calculating the length of a barrel or rails?

The formula for calculating the length of a barrel or rails is: L = (m*v)/(F*tanθ), where L is the length, m is the mass of the projectile, v is the desired velocity, F is the force applied by the propellant, and θ is the angle of the barrel or rails.

How does the angle of the barrel or rails affect the length calculation?

The angle of the barrel or rails (θ) is an important factor in the length calculation because it determines the direction of the force applied by the propellant. A larger angle will result in a longer length, as the force needs to act over a larger distance to achieve the desired velocity.

Are there any other factors that need to be considered when calculating the length of a barrel or rails?

Yes, in addition to the factors mentioned above, you may also need to consider factors such as air resistance, friction, and the shape of the projectile. These can affect the force and velocity of the projectile, and thus impact the length calculation. It is important to account for all relevant factors for accurate results.

Is there a standard length for the barrel or rails in basic Newtonian space physics?

There is no standard length for the barrel or rails in basic Newtonian space physics, as it will depend on the specific parameters of your experiment or application. The length will vary depending on the desired velocity, projectile mass, and other factors. It is important to calculate the length based on your specific needs.

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