Hovertanks

[Lancefighter]

Why do I bother?

A jet engine is classified as an engine that uses air as the moving part of the engine. The gas reacts.. and well, explodes, pushing air backwards. This is either direct thrust, as in lighter craft, or turning a turbine for heavier stuff.

An internal combustion engine is any engine where the combustion happens inside the engine. All jet engines are internal combustion. The only real external combustion engine is a steam engine, and well, we dont use those much.

A 'reaction engine' is anything where the components react. Well, this is true in all combustion engines. Why? Because combustion is a reaction.

Now, if you would please stop saying things that make no sense? your using a good deal of big words, but your not actually saying anything.

[TVR]

Do you even understand why turbo-shafts would work, but turbofan, turbojet, and turboprop wouldn't?

[Lancefighter]

As an aerospace engineering student, I know how all of these work. You are not saying anything that makes sense - There is no reason a turboshaft would work that the other jet engine types wouldnt. I realize the major difference is how the force is applied (turning something instead of pure exhaust thrust)

As such, I understand that yes, because the force is applied differently, it is more useful in tanks, helicopters, etc.

[lav_coyote25]

Do you even understand why turbo-shafts would work, but turbofan, turbojet, and turboprop wouldn't?

use the force TVR, use the force.

[TVR]

Do you believe that moving the intake from the side to the front of the propulsion unit would cause no problems?

[Lancefighter]

Not sure what your asking. Because, well........
The jets turn a propeller. The air is both taken from the side and the front of the prop..

[TVR]

Here's a front view of the LCAC, care to illustrate where the 'front intake' is?

... A 'reaction engine' is anything where the components react. ...

A reaction engine is an engine that creates linear acceleration by expelling propellant as dictated by Newton's Third Law of Motion, you should already know why turboshaft and piston engines aren't reaction engines.

(Hint: They don't create linear acceleration by action-reaction, they only create ANGULAR motion.)

[Lancefighter]

I said there was an intake in front of the prop, that the prop uses to push air.. was that not evident?

[Zarel]

Newton's Third Law states that the force (ie, the Kinetic Energy) must be equal, otherwise it would violate the conservation of momentum.

I haven't been participating in this discussion for fear of getting something wrong, but I'm pretty sure force and kinetic energy are not the same thing... Newton's Third Law is about how net force is always zero. Conservation of energy is something else entirely, and kinetic energy is rarely conserved - it's usually dissipated into other forms of energy.

[3drts]

Exactly, he doesn't seem to realize that kinetic energy is not the same as momentum.
Momentum varies linearly with velocity, kinetic energy increases exponentially, with the square of velocity.
Thus is is quite easy to have two objects pushed apart, with the force pushing them apart exactly equal, but one of the objects departing with much much much more energy - such as when a gun fires.
The gun recoils backward, but the KE of the gun is rather low.
Consider a 5.56*45mm NATO round - a 55 grain (3.5 gram- 0.0035 kg) projectile traveling 990 m/s.
its kinetic energy is 1/2 0.0035 * 990^2 = 0.0035 * 980100 = 1/2* 3,430 joules. 1,715 joules

By comparison, an M-16 is about 2.2 kg, to have that much KE, the thing would need to recoil at a speed of
(1715/2.2)^(0.5) = 28 m/s - about 100 kilometers per hour.
If that were the case, by firing such a weapon downward, without holding it (say you just put the muzzle on the ground, balance it/support its sides, and push the trigger with a stick from below), the thing would go flying up for about 3 seconds, then fall down, for a total airtime of 6 seconds, and a max height of 45m = about 150 feet.
Such doesn't happen.
You need to brush up on your physics.

Lance fighter- those fans are not intakes to the turbine.

Air just passes through them, as fan blades push the air back.
Those fan blades are connected only by drive shafts to the turbine shaft, which must take air in somehow, to burn, and pass through the turbine blades.

But as I showed with the XF-2Y Sea Dart, and other flying boats, Turbojets (jet turbines, that propel a craft principally with the jet stream, and not mechanically by something attached to the shaft) can be placed on waterborne craft.
The problems are not insurmountable - the craft and placement of intakes is meant to reduc

[Rman Virgil]

.

This discussion reminded me of the SRN4 which operated for almost 40 years between Dover and Calais (over water), beginning in the early 60's.

It was capable of carrying over 380 passengers and 40 cars, was was 91 feet wide,185 feet in length, weighed 300 tons, could cruise at 70 mph and had props 20 feet in diameter.

With all that in mind, could there be any logical weaponized version ? Over water, some ideas occurred that seem sensible.

Over land ? Too much working against it to make sense, realistically. Stuff already mentioned and things like being every bit as vulnerable to anti-tank land mines as trax or wheels.

Though, hover tanks in com-vid games have been very popular since first introduced as such back in 1991 I think. Enduring for well nigh 20 years as game entertainment does make for a strong impression.

- RV
.

[Lancefighter]

Disclaimer - At this point im arguing for the sake of arguing, not really for useful reasons :p

Yousee, the air is taken in on one side of the prop duct, and propelled out the back. So, well, its an intake on the front of the duct.

[TVR]

... Momentum varies linearly with velocity ...

I've been thinking about this, and it would imply the force experienced during deceleration would be the exact same, minus some loss due to conversion of energy, as during the acceleration.

As the launcher, projectile, and target form a closed system, which is required for to determine projectile velocity, this means the target of similar mass would experience a linear transfer of momentum equal to that of the launching platform.

Say that the 5.56 NATO bullet exerts 7 joules of recoil on the shooter, wouldn't this mean that a target of similar mass would only need 7 joules to decelerate that bullet?

[Lancefighter]

Yes, but the problem is that a bullet uses the whole pressure equation - because the bullettip is thin, it punches through stuff leaving less drag as it goes through stuff. So sure, 7 joules, but the problem is thats actually 7 joules APPLIED to the bullet, and because of its aerodynamics its not as easy as it seems.
Oh, and usually punching a hole in things kills stuff, so the problem isnt stopping the bullet, the problem is stopping the bullet before it goes through someone :p

[3drts]

You are both wrong.

Lets be clear, 7 joules is not a valid recoil value.
Recoil may be measured in units of momentum, mass * velocity, such as kg*m/s

A joule is a product of force * distance, which is mass*acceleration*distance
http://en.wikipedia.org/wiki/Joule
(kg*m^2) /(s^2)

Now, a rifle bullet is typically accelerated over a distance of 20-24 inches (508 to 620 mm)
To stop the bullet in that same distance, the same average force (averaged) will be needed.
Now, this makes perfect sense with why the bullet has more energy than the gun- though the exact same force is applied at the exact same time, the force on the bullet is applied over a distance of about half a meter, whereas the gun+shooter system barely moves at all during the time it takes the bullet to leave the barrel.

And before you ask about why distance matters,
Consider when you depress an object on a spring, or push an item up a hill, the distance you move it corresponds to the energy you've given it
- in the simple case of a hill, you increased its potential energy, and the force required to do so is linear with height.
This should be fairly obvious.
And you will find this conversion of PE to KE follows m*g*h = 1/2 m*v^2 (which can be derived from the distance equation of 1/2 a*t^2, which itself is the derivative with respect to dv/dt of the distance equation that is simply v*t - basic calculus, distance being a function of velocity*time should also be fairly obvious)

an M-16 firing "in midair" (lets say a 3 shot burst is fired, and the gun is dropped after the first shot, but before the third, so the gun recoils freely - we can ignore shooter mass and such), may have a KE of 7 joules as a result of the recoil (I'm not actually going to calculate this),

When that bullet hits, and stops in a more massive target, the same momentum will be transfered to the target, but the resulting "recoil" of the target will result in a KE with far less than the 7 joules the gun firing in midair had.

The energy of the bullet (1,000 + joules, assuming little air resistance, assuming it is stopped by the target) is completely dumped into the target.
That doesn't mean all the bullet's KE is transfered to the targets KE.
There are shock-waves and such throughout the targets insides, etc, and all the energy is eventually turned to heat.

Its also worth noting that most of the time, the bullet penetrates the target, and keeps going and keeps most of its KE to itself.

Say that the 5.56 NATO bullet exerts 7 joules of recoil on the shooter, wouldn't this mean that a target of similar mass would only need 7 joules to decelerate that bullet?

No, the target will need to exert a force, over a distance, that is equal to the bullets KE.
The result of exerting this force on the bullet, will transfer momentum to the target, thereby giving the target some KE, but this can be any value depending on the targets mass.

the air is taken in on one side of the prop duct, and propelled out the back. So, well, its an intake on the front of the duct.

The problem is this is completely irrelevant. Where the prop gets its air is of no consequence to where the turbine gets its air, and thats what you guys were arguing about.
The prop is not the jet turbine. None of the air that goes in the prop turbine goes to the turbine.