Missile energy (split)

[EtherealN]

No, it should be a lot more. (See if I can get it right, waaaay too late to do maths.)

 

KE = ½ * M * v²

 

½ * 136 * 700²

 

½ * 136 * 490 000

 

136 * 245 000

 

KE = 33 320 000 Joules

 

Note, this is for 700 meters per second. Saying what the value is for "mach 2" is impossible - mach 2 in what atmosphere/temperature etcetera?

 

EDIT: let's see, the 700m/s should be slightly above mach2 at 30k feet standard atmosphere (-44 degrees C). Almost exactly mach2 at sea level, standard atmosphere (15 degrees C).

Edited August 11, 2012 by EtherealN

[Exorcet]

You can get speed of sound from sqrt(gamma*R*T). Assume gamma = 1.4, R = 287, T is in degrees Kelvin.

 

T as a function of height can be approximated by

 

216.65 + 2*ln[1 + exp(.5*{sea level Temp - 216.65 - 6.5*y})]

 

y is altitude in kilometers.

 

 

If you want to model a missile in flight, you can do it through energy balance. Rocket engine power output can integrated over to get the kinetic energy of the missile and the total amount of fuel burned. Drag is converted to energy by multiplying by velocity and this is simply subtracted from the rocket power to determine the net power acting on the missile. The tricky part is finding accurate values for real missiles.

[mvsgas]

No, it should be a lot more. (See if I can get it right, waaaay too late to do maths.)

 

KE = ½ * M * v²

 

½ * 136 * 700²

 

½ * 136 * 490 000

 

136 * 245 000

 

KE = 33 320 000 Joules

 

Note, this is for 700 meters per second. Saying what the value is for "mach 2" is impossible - mach 2 in what atmosphere/temperature etcetera?

 

EDIT: let's see, the 700m/s should be slightly above mach2 at 30k feet standard atmosphere (-44 degrees C). Almost exactly mach2 at sea level, standard atmosphere (15 degrees C).

I thought mach two was 680 meters per second 462400. I guess we need a specific altitude ha

[EtherealN]

Exorcet gave it to you. :)

[combatace]

So how do you get the mass of the vehicle? It is not just the weight IIUC

 

If missile x weights 136 kilos, what is its mass?

 

If you talk mass 136kg is its mass, weight is actually mass x gravitational acceleration and its units is kg.meter/second^2.

[SgtPappy]

Which is simplified as Newtons, a unit of force. I like how we're keeping it at SI units!

[PeterP]

Did you know: I love the Metric system!

 

Epic thread! Thanks!:) - This goes directly to my bookmarks.

 

 

BTW: how many Mach have 20 spoons and a quarter elbow of cups at 15legs while moving with 3 knots in the pocket trough a very dense air?

 

Edit: Oh-oh ! - I forgot: The 20 spoons have a mass of 3 pounds on each kilogram.

so once again: How many Mach? - I hope its not much because I can carry only two 15liters plastic bags with me...

 

 

I think that would be appropriate, enough people hate me already.

 

no way! thanks for your input!- remember : there isn't sometime like a "stupid" question - it's always the answer that let it look stupid! :)

Edited August 18, 2012 by PeterP

[EtherealN]

Can't help you with that PeterP, but I can tell you that if you eat a Quarter Pounder with cheese, your body will gain roughly the same amount of energy as a 136kg missile has stored in kinetic energy when traveling at 20 km/h.

 

:P

[PeterP]

but I can tell you that if you eat a Quarter Pounder with cheese, [...]

:P

 

Forward or backward?! :P

[EtherealN]

Both. At the same time.

[PeterP]

Thanks!

Seems it only needs 118.343.200 Flies to crunch a thread. BTW : a average Fly weights about 0,1352 gram...

Edited August 12, 2012 by PeterP

[Teapot]

PeterP .. is that what 16 New Tons looks like?

BTW, when it's inert (just sitting there) it's not a unit of force, but of potential force.

If one were to drop it onto one's head from ceiling height (for arguments sake) ... only then would it be a unit of force. Well, 16 new tons of it :smartass:

Edited August 12, 2012 by Teapot

[Corrigan]

BTW, when it's inert (just sitting there) it's not a unit of force, but of potential force.

If one were to drop it onto one's head from ceiling height (for arguments sake) ... only then would it be a unit of force. Well, 16 new tons of it :smartass:

 

Well... I've never heard of that concept. Either a force is exerted, or it is not. ;)

Edited August 12, 2012 by Corrigan

[Kuky]

What will happen in the future is missiles that never miss.

 

In the future?? I can build you a missile like that right now.. but no one will buy it - who wants a missile that flies slower than a plane anyway..

 

By "what will happen in the future" I meant the thrust power of a launching aircraft, if launching aircratf is powerfull enough it potentially could exceed the missile max speed, take SR-71 or MiG-25 as example, both can fly above mach 3, so who knows what kind of propulsion aircraft and missiles will have adn who knows how things will change in the future... physics won't change though :smartass:

[asparagin]

a missile that flies slower than a plane anyway..

 

if launching aircratf is powerfull enough it potentially could exceed the missile max speed

 

I see no difference.. :smartass:

 

(I understood what you meant from the beginning..)

[marcos]

Take a 120, which can hit maybe 3.3M at altitude from a 0.9M launch. Launch it at 2.0M and it will hit 4.5M.

As for statements like 'the kinetic energy was the same', I throw the book of physics at you :P

Surely the KE difference between M2.0 and M4.5 is way larger than the KE difference between M0.9 and M3.3?

 

What will happen in the future is missiles that never miss.

Wouldn't bet on that. Once the pilot is removed from the equation, the artificial 9g limit will also be removed, then you'll have a battle of superior thrust-to-weight vs superior lift, and microchip vs microchip.

Edited August 12, 2012 by marcos

[EtherealN]

so who knows what kind of propulsion aircraft and missiles will have adn who knows how things will change in the future...

 

Ramjets. Already happening. :)

[EtherealN]

Surely the KE difference between M2.0 and M4.5 is way larger than the KE difference between M0.9 and M3.3?

 

Yes. Assuming the 136 kilogramme rocket:

 

M0,9 = 4 957 200 Joule

M2.0 = 24 480 000 Joule

M3.3 = 66 646 800 Joule

M4.5 = 123 930 000 Joule

 

BUT, this is irrelevant. It's not the KE that gets the missile FROM the aircraft's speed to it's top speed. A bomb carried on the same aircraft, with a 136kg mass, would have the same KE.

 

The difference is that up high, there is less air. This means less drag. This means not only one, but two things:

 

1) Higher top speed.

2) More acceleration.

 

That is, at any given speed, you will have less drag when at altitude compared to when at sea level. This means the rocket motor has less drag to overcome. This means more impulse left after drag is "deducted". So you'll accelerate faster, and in the example GG was making - your acceleration is happening from an already fast position! Remember, the KE here only discussed the energy the missiles have before igniting the rocket motors! The rocket motors will then begin to convert more potential energy (a.k.a "fuel) into even more Kinetic Energy. ;)

 

(And some of it into heat and so on.)

[marcos]

Yes. Assuming the 136 kilogramme rocket:

 

M0,9 = 4 957 200 Joule

M2.0 = 24 480 000 Joule

M3.3 = 66 646 800 Joule

M4.5 = 123 930 000 Joule

 

BUT, this is irrelevant. It's not the KE that gets the missile FROM the aircraft's speed to it's top speed. A bomb carried on the same aircraft, with a 136kg mass, would have the same KE.

 

The difference is that up high, there is less air. This means less drag. This means not only one, but two things:

 

1) Higher top speed.

2) More acceleration.

Appreciated but the OP didn't specifiy a different altitude.

 

The missile will be accelerating when its fuel runs out. The faster it's going, the less it accelerates (because drag force increases as speed increases), but it will still accelerate beyond the speed it would reach if you launch it at M0.9. Take a 120, which can hit maybe 3.3M at altitude from a 0.9M launch. Launch it at 2.0M and it will hit 4.5M.

 

So really you've just got thrust vs drag and the drag is far greater at higher speeds (0.5*Cd*Density*A*V^2), unless the difference can be explained by the huge increase in Cd within the transonic regime. Wouldn't have thought that would be of the same order of magnitude though.

 

That is, at any given speed, you will have less drag when at altitude compared to when at sea level. This means the rocket motor has less drag to overcome. This means more impulse left after drag is "deducted". So you'll accelerate faster, and in the example GG was making - your acceleration is happening from an already fast position! Remember, the KE here only discussed the energy the missiles have before igniting the rocket motors! The rocket motors will then begin to convert more potential energy (a.k.a "fuel) into even more Kinetic Energy. ;)

 

(And some of it into heat and so on.)

Fully realise that but although I haven't actually done the calcs I'd be surprised if the fuel required to get from M=0.9 to M=3.3 is the same as that required to get from M=2.0 to M=4.5 at the same altitude. I also thought potential energy is the energy due to a mass being at altitude in a gravitaional field. Fuel is stored chemical energy held within the bonds of the propellant chemical.

Edited August 12, 2012 by marcos

[mvsgas]

Exorcet gave it to you. :)

:doh::D

That is mart people talk, me not know it:joystick:

 

So, since me not smart used the power of the Google

IIUC mach at 30, 000 feet (9144 meters) is 303.1 meter per second (not sure if it maters but 229 kelvin)

 

KE=0.5 x 136 x 303 squared

KE=0.5 x 136 x 91809

KE=136 x 45904.5

KE=6243012

 

is that closed?

 

What is the thrust we are giving this rocket?

How much the the fuel weight? I think this matters since it will be lighter or less massive right?

How long will the engine burn? 20 seconds?

 

 

Edit

D'oh I think that is mach one. Domo stupido Google! D'oooohh, hold on

Edited August 12, 2012 by mvsgas

[EtherealN]

marcos, GG was illustrating a point, not giving you the absolute and correct numbers for any given missile. ;)

[EtherealN]

What is the thrust we are giving this rocket?

How much the the fuel weight? I think this matters since it will be lighter or less massive right?

How long will the engine burn? 20 seconds?

 

All of these are irrelevant for the problem at hand. What was sought to show was that you will have more energy to start with, and as long as you don't force the missile to fly the entire trajectory at sea level, there's just no way the motors will get you to thrust=drag. Thus, since we don't have that "top" level relevant, it's fair to assume that if you have two cases with different amounts of energy at start, and give them the same amount after that, the one that started with more will have more at motor burn out.

 

I'll warn you, if you want to do a full computation of a missile flight, the mathematics will get WAY more complex than what has already been touched on so far. ;)

Especially if we want to to an anywhere close to accurate-for-modern-missiles flight profile.

Edited August 12, 2012 by EtherealN

[mvsgas]

Ok screw it lets say it is 607 meters per second squared

KE=24918668J ?

 

Domo stupido math/algebra. Stupid bad "edumacation"/little brain...alcohol does kill brain cells...stupid Duff beer

[mvsgas]

All of these are irrelevant for the problem at hand. What was sought to show was that you will have more energy to start with, and as long as you don't force the missile to fly the entire trajectory at sea level, there's just no way the motors will get you to thrust=drag. Thus, since we don't have that "top" level relevant, it's fair to assume that if you have two cases with different amounts of energy at start, and give them the same amount after that, the one that started with more will have more at motor burn out.

 

I'll warn you, if you want to do a full computation of a missile flight, the mathematics will get WAY more complex than what has already been touched on so far. ;)

Especially if we want to to an anywhere close to accurate-for-modern-missiles flight profile.

 

No the problems was that initial speed affected terminal one. But needed to get the initial Jules.

[EtherealN]

Ok screw it lets say it is 607 meters per second squared

KE=24918668J ?

 

Domo stupido math/algebra. Stupid bad "edumacation"/little brain...alcohol does kill brain cells...stupid Duff beer

 

I get 25 054 532 Joule.

 

½ * 136 * 607²

 

½ * 136 * 368 449

 

136 * 184 224,5

 

25 054 532