Physics Question by AnthonyT

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Physics Question

POSTED BY: ANTHONYT
UPDATED: Tuesday, February 13, 2007 04:31
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Friday, February 9, 2007 4:20 PM

Freedom is Important because People are Important

Hi all,

I could use some basic science help.

I would like to know how to calculate the amount of force needed to launch a given weight a given distance, and I'd like to know how fast it will travel throughout the duration.

Essentially, I guess whatever the calculation is that artillery men use.

Can anyone help?

--Anthony

P.S. Please only people who believe that there is visible light on the moon reply.

"Liberty must not be purchased at the cost of Humanity." --Captain Robert Henner

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Friday, February 9, 2007 4:52 PM

6IXSTRINGJACK

Quote:
Originally posted by AnthonyT:
P.S. Please only people who believe that there is visible light on the moon reply.

Hahahahahahahaha......

"A government is a body of people, usually notably ungoverned." http://www.myspace.com/6ixstringjack

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Friday, February 9, 2007 5:04 PM

RUE

I have a vote and I'm not afraid to use it!

It's a simple problem but it takes a lot of translation and you have to make some assumptions. One is that you'll aim your launcher at 45 degrees to get maximum distance. The other is that you're looking for the equation on earth as the accelaration due to gravity is important in the equation. Without these assumptions your equation is full of intimidating symbols.

And I just found out I have to go. I'll get back tomorrow (I hope) with the equation with the two assumptions.

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Friday, February 9, 2007 5:18 PM

6IXSTRINGJACK

Nothin' like a little suspense, huh.

"A government is a body of people, usually notably ungoverned." http://www.myspace.com/6ixstringjack

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Friday, February 9, 2007 5:36 PM

FINN MAC CUMHAL

The kinds of calculations that you’re referring to are not trivial, and typically require sophisticated algorithms to solve external ballistic equations of motion, incorporating everything from spin-stabilization to hypersonic inviscid flow fields. The answer depends on the depth that you want to get into the subject. On an introductory level, there are simple formulas of physics that can be applied but while these formulas are correct in theory they aren’t sufficient to describe flight path or energy of modern munitions. As such most artillerymen don’t actually perform these kinds of calculations. In many cases the field gun sighting is left up to computers or charts calibrated to the gun or missile.

To try to provide some answer to your question, the force that acts on an artillery shell is called an impulse, and calculating it exactly is really beyond the scope of this forum, since it requires an understanding of the thermophysics of the charge, the compressibility and energy of the resultant gases and the elasticity of the projectile. (An entirely separate subject called internal ballistics.) But if you know the distance that the projectile travels and the mass of the projectile, then the impulse can be calculated using the formula:

I = M * SQRT( ( R*g ) / sin (2*A) )

Where M is the mass of the projectile, R is the range, g is the acceleration of gravity (ass. g=9.8 m/s^2) and A is the launch angle.

To know the force involved, you must know how long the shell was in the barrel; that will give the dt. The force then is:

F = I / dt

Once the munition is airborne it becomes a matter of ballistic motion, and once again making many simplifying assumptions, the velocity can be understood as composed of two components, a horizontal and a vertical component, with the chief difference between the two being that the vertical component will be effected by gravity.

V_horizontal = V_initial * cos ( A )

V_vertical = V_initial * sin (A) – g * t

V_initial can be found with:

V_initial = SQRT ( (R*g) / sin (2*A) )

Where A is again the launch angle, g the acceleration of gravity and t the time into flight.

If you’re interested I can give you sources that provide a more advanced treatment.

Nihil est incertius vulgo, nihil obscurius voluntate hominum, nihil fallacius ratione tota comitiorum.

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Friday, February 9, 2007 8:02 PM

ANTHONYT

Freedom is Important because People are Important

Wow,

I am beginning to wonder if this math isn't beyond me.

Would it be easier if we were using a catapult?

--Anthony

"Liberty must not be purchased at the cost of Humanity." --Captain Robert Henner

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Friday, February 9, 2007 8:15 PM

6IXSTRINGJACK

Quote:
Originally posted by AnthonyT:
I am beginning to wonder if this math isn't beyond me.

Would it be easier if we were using a catapult?

I recommend playing "King Arthur's World" on Super Nintendo. You can build your own catapults and adjust the fire to hit just about anything.... even Arthur, which I wouldn't recommend.

"A government is a body of people, usually notably ungoverned." http://www.myspace.com/6ixstringjack

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Friday, February 9, 2007 8:22 PM

CHRISISALL

Quote:
Originally posted by Finn mac Cumhal:

If you’re interested I can give you sources that provide a more advanced treatment.

"I only need to know one thing; where...they...are."

Too bad Chrisisall

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Saturday, February 10, 2007 1:53 AM

CITIZEN

Quote:
Originally posted by chrisisall:
"I only need to know one thing; where...they...are."

Too bad Chrisisall

They said alien but Chrisisall thought they said ILLEGAL aliens and signed up.

HudsonIsAll.

More insane ramblings by the people who brought you beeeer milkshakes!
No one can see their reflection in running water. It is only in still water that we can see.

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Saturday, February 10, 2007 2:54 AM

GEEZER

Keep the Shiny side up

As well as the stuff Finn so ably described, don't forget external ballistics. All sorts of computations based on: the shape, weight, and density of the projectile; whether it's tumbling or stabalized; air density, temperature, altitude, humidity, and wind.

There are plenty of free external ballistics programs on-line that would give you speed at distance, but they're mostly designed for long range hunting or target applications. You'd have to know at least the coefficent of drag (more formulas for that), weight, and launching velocity of your projectile.

"Keep the Shiny side up"

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Saturday, February 10, 2007 7:53 AM

SIGMANUNKI

Google'd projectile motion and found:

http://library.thinkquest.org/2779/Even_more.html

http://id.mind.net/~zona/mstm/physics/mechanics/curvedMotion/projectil
eMotion/generalSolution/generalSolution.html

Which is high-school physics.

If you want all the theory, check out a physics text from your local library. Something like "Physics for Scientists and Engineers" published by Brooks/Cole would be good (I think current is 6th ed). It's the standard 1st year text, but I refer to it as a picture book. As in, it re-does high-school and then continues. But, from what you have posted, I doubt that you'll need the continuation.

If you just want the bare bones "Physics for Game Developers" by O'Reilly is an excellent treatment.

Hope that helps

----
I am on The List. We are The Forsaken and we aim to burn!
"We don't fear the reaper"

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Saturday, February 10, 2007 9:24 AM

FINN MAC CUMHAL

Quote:
Originally posted by AnthonyT:
Wow,

I am beginning to wonder if this math isn't beyond me.

Would it be easier if we were using a catapult?

I don’t know. That depends a lot on what you’re trying to do. Are you assaulting the beaches of Normandy or besieging the walls of Orléans?

There’s no doubt that the field of ballistics can challenge almost anyone’s mathematical knowledge, but like I said it depends on the depth that you want to get into the subject. You won’t be able to calculate the trajectory of modern munitions, in general, without a fairly complete treatment, but if you have some flexibility to apply some basic simplifying assumptions, the complexity of the math can be reduced dramatically.

One assumption is to treat your projectile as a point mass. That will eliminate, for the most part, any discussion of spatial orientation. Real projectiles can have many degrees of freedom. They can spin, precess, yaw and do all kinds of crazy ballet in their flight dynamics. Eliminating that will free you from keeping up with things like angular velocities and static margin. This is a fair assumption for a smooth spherical projectile. It’s not a good assumption for less spherical shapes, like bullets, or for dimpled spheres, like golf balls. In those cases you need to understand orientation; the dimples on spinning golf ball, for instance, can cause the ball to curve way out of trajectory plane, requiring the need for a lateral component.

Another simplifying assumption is to treat the projectile as slow moving. Since drag goes directly as velocity, the faster your projectile moves the more drag will affect your trajectories. A slow projectile might reduce drag enough to be tolerable, and that will eliminate the need for things like ballistic coefficients, if your targeting is flexible enough to allow for some trial and error. The ballistic coefficient is really not a constant; it goes as the velocity projected area, so a spinning odd shaped projectile could have a ballistic coefficient that various rather dramatically and randomly. If the above point-mass assumption is assumed, then the ballistic coefficient is undefined anyway, since the projected area of a point-mass is always zero. At high velocities, like those associated with bullets, the drag effects are significant even for the relatively small velocity projected area. And at hypersonic velocities, like those associated with ballistic missiles, the drag effects become so dominant that they can rip the projectile apart.

But the drag force can be estimated for slow moving objects using the formula:

Fd = 0.5*Cd*rho*A*v^2

Where Cd is the drag coefficient that can be thought as describing the shape (which for a sphere could be a high as 0.5), rho is the density of air (1.115 kg/m^3), A is the velocity projected area, and v is the velocity.

As you can see the drag force increases as the square of the velocity.

Alternatively, and generally in practice, computers are used to do these kinds of calculations, so Geezer’s approach of searching for an online ballistics tool could be a good idea if you can't apply the above simplifying assumptions.

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Saturday, February 10, 2007 9:32 AM

RUE

I have a vote and I'm not afraid to use it!

Well, it looks like all the answers are here already - from the simple HS physics version (which I was going to supply) to the somewhat advanced ones.

I guess I'll step out now.

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Monday, February 12, 2007 8:14 PM

PIRATENEWS

John Lee, conspiracy therapist at Hollywood award-winner History Channel-mocked SNL-spoofed PirateNew.org wooHOO!!!!!!

Quote:
Originally posted by AnthonyT:

I would like to know how to calculate the amount of force needed to launch a given weight a given distance, and I'd like to know how fast it will travel throughout the duration.

Essentially, I guess whatever the calculation is that artillery men use.

Can anyone help?

--Anthony

FYI: http://en.wikipedia.org/wiki/Rifleman's_rule
http://www.biggameinfo.com/BalCalc.aspx
http://www.handloads.com/calc/

Quote:

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Tuesday, February 13, 2007 4:03 AM

AURAPTOR

America loves a winner!

Quote:
Originally posted by AnthonyT:
Wow,

I am beginning to wonder if this math isn't beyond me.

Would it be easier if we were using a catapult?

--Anthony

"Liberty must not be purchased at the cost of Humanity." --Captain Robert Henner

I suppose that depends on what you're trying to launch and how far. If you're gonna go siege engines, I'd suggest a Trebuchet.

People love a happy ending. So every episode, I will explain once again that I don't like people. And then Mal will shoot someone. Someone we like. And their puppy. - Joss

" They don't like it when you shoot at 'em. I worked that out myself. "

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Tuesday, February 13, 2007 4:31 AM

FINN MAC CUMHAL

I built a Roman ballista for an American Way charity event to launch tennis balls at a bucket of water. The idea was to get people to buy ‘shots’ until the bucket was tipped over. It was five dollars a tennis ball and I ended raising something like $500 for charity.

It had a tremendous amount of power. In fact, I ended up cracking one of the arms and had to stop. But it was horribly inaccurate. Oddly, the Romans didn’t design it to shoot tennis balls. If could hit the bucket I could push it a good distance to the edge of the table, but I never managed to knock if off the table, because only like one out of maybe a dozen shots actually hit the bucket. When the arm cracked it was too dangerous to continue using so I had to give up.

Here’s a discussion of the design of the Roman Ballista. The one that I build was a modern modification of the Heron cheiroballista, which was the most advanced dual torsion artillery used by the Romans.
http://www.unc.edu/courses/rometech/public/content/special/Kat_Smith/R
OMANARTILLERY.htm

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