HP vs Torque vs Fuel Efficiency

[niky]

P=mv only
F=ma.

As I recall, there is no term that describes mva. A is already delta-v, so you're just doubling it up needlessly.

[jakobnev]

If you use a microwave engine what would you be measuring your speed against then?

A padded room.

[mort]

Originally Posted by darcane

Nonsense?

Jet engines are rated in Thrust (which is really just a Force) rather than Power. With a constant thrust, power increases proportionally to speed. If you eliminate aerodynamic drag by being in space, then Acceleration is constant if Thrust is constant.

This is why he specified a rocket engine (which is a type of jet engine).

Hi darcane,
Jets, and of course rocket engines are specified at least 2 levels of thrust. Sea level at 0 speed; and maximum. If the engine is for the military one or more contract requirements will also be published. The maximum thrust occurs at exactly one altitude and speed. Above or below either and the thrust will be less. Sometimes there will also be a chart of thrust vs speed at constant fuel burn (that's at the maximum power level) and constant altitude. That chart generally shows a slight increase in thrust below some point, because the engine might make more power as it runs leaner. Then the thrust declines just faster than the speed increases. True for most rockets that specify thrust also - it is the maximum available, often the altitude and speed are also given.
For rocketry the specific impulse is often specified. In the wikipedia article you will find this interesting statement:

For rockets and rocket-like engines such as ion-drives a higher I_{sp} implies lower energy efficiency: the power needed to run the engine is simply:

dm/dt * v_e^2/2/

where ve is the actual jet velocity.

whereas from momentum considerations the thrust generated is:

dm/dt * v_e

Dividing the power by the thrust to obtain the specific power requirements we get:

v_e/2

Hence the power needed is proportional to the exhaust velocity, with higher velocities needing higher power for the same thrust, causing less energy efficiency per unit thrust.

-mort

[mort]

Originally Posted by niky

I did. Twice now. I am still puzzled why you think that it actually disproves my point. Unless you're suggesting that a rocket and its exhaust form a non-inertial frame of reference... because they don't.

Hi niky,
First: I am saying the rocket accelerating is not in an inertial reference frame. If you hold a ball and let it go in side the rocket, it falls as if acted on by an invisible force. If you toss the ball up it follows a parabolic path, not the straight line required by F = ma. But the wikipedia article does say that by adding a "fictional" force (I hate that term) - in this case graviticity (like gravity) you can make a transform that lets you use an inertial frame of reference.
Second: you had said:

In free space, there are no other inertial frames to measure exhaust velocity by.

This is not true. One point made in the wikipedia piece is that for any 2 inertial frames of reference (IFoR from now on) a Galilean transform can be found to relate the displacement and time of the 2 IFoRs. That means in one frame I can find a transform that allows me to measure the speed of an object in any other IFoR.
Not knowing the speed inside one IFoR is not the same as not having a speed. If the rocket is accelerating what you don't know is the initial speed. You can easily compute the additional speed due to the acceleration - this is the speed I care about. And as this speed increases geometrically increasing power to accelerate a given mass at a constant rate is required.
The fact that a legitimate IFoR can be found in which the speed compared to some other object is unknown isn't a feature. If there is some object, another rocket for instance, intersecting with the path of my rocket, I could look out the window and measure our respective velocities, thus expanding the space of interest in my IFoR. Or I could claim that I don't know my speed. When my rocket hits the other rocket then I suddenly know the relative speeds by measuring the sudden deceleraBANG.
The IFoR can always be expanded to include objects you are getting nearer to or farther from - and hence you can always know your speed.
In the case of the train and bullet, which I don't see why you thought that was interesting, people on the train know it is going 1000 mph, people off the train do too. I'll allow that the gun and bullet don't know, but the observer off the train measures the speed of the bullet and can make the simple transform to know what relative velocity the bullet leaves the gun. Nothing there is inconsistent with what I have been saying all along.

But this:

Of course, the inventor claims that the radiation never leaves the device, and that it can produce electricity on deceleration (in spaaaaaace), which means that he has just invented perpetual motion. Since deceleration is merely acceleration in the opposite direction.

is stupid. Does regenerative braking imply perpetual motion? From looking at the Maxwell's equations basis of the theory it looks like their power efficiency is below 10%
Edit - I just reread the original em drive paper, and they claim that with wave guide Q = 50,000 which is currently available, efficiency is 60% (I had mistaken it as 6%) - remember this is not the efficiency of the engine, but more like the gearbox.
-mort

[sendler]

Originally Posted by mort

And as this speed increases, geometrically increasing power to accelerate a given mass at a constant rate is required.

So let's say we take the motorcycle drive train away and replace it with a ducted fan which is geared to run best at the power peak of the engine and has the same efficiency throughout the speed range of the test. Discounting air resistance and any losses on the vehicle, and holding 100 hp constant from the engine. Will we have constant values for thrust and for acceleration to the IFoR of the rider?

[mort]

Originally Posted by darcane

However, your example is still misleading because the effects of aerodynamic drag increase exponentially and very quickly dwarf the power needed for acceleration.

Hi darcane,
OK, misleading. I was going to have a chart showing power for RR cdA and acceleration all vs speed with limits and stuff. It is kind of a mess, here it is. This is for a 600 lb bike with A=5 sq ft and cdA= 0.5 - assumptions: 100 hp available from engine, excess hp goes to accelerate the bike - which is crazy at 10 mph it only takes 16 hp for 1 g.

speed HP  Aero  RR  Accel   hp for   [accel using
MPH total  hp   hp  hp avail 1 g      avail hp]
 10  100     0  0.2  99.8      16      6.2
 20  100     0  0.3  99.5      32      3.1
 30  100     0  0.5  99.1      48      2.1
 40  100     1  0.6  98.3      64      1.5
 50  100     2  0.8  97.1      80      1.2
 60  100     4  1.0  95.3      96      0.99
 70  100     6  1.1  93.0              0.83
 80  100     9  1.3  90.0              0.70
 90  100    12  1.4  86.1              0.60
100  100    17  1.6  81.3              0.51
110  100    23  1.8  75.5              0.43           
120  100    30  1.9  68.5              0.36
130  100    38  2.1  60.4              0.29
140  100    47  2.2  50.8              0.23 <-- aero and accel hp about equal
150  100    58  2.4  39.9              0.17
160  100    70  2.6  27.4              0.11
170  100    84  2.7  13.3              0.05
180  103   100  2.9  0.0 <-- top speed

So at 100 mph you have 81 hp available to use to accelerate. Which allows about .5 g
-mort

[mort]

Originally Posted by sendler

So let's say we take the motorcycle drive train away and replace it with a ducted fan which is geared to run best at the power peak of the engine and has the same efficiency throughout the speed range of the test. Discounting air resistance and any losses on the vehicle, and holding 100 hp constant from the engine. Will we have constant values for thrust and for acceleration to the IFoR of the rider?

Hi sendler,
No.
Thrust and acceleration measure the same thing. When thrust = the weight of the bike the acceleration is 1 g. As the speed increases the thrust will fall.
This is almost an airplane propeller for which the basic formula reduces to this:
Thrust = ( HP * eff * 375 ) / air speed (mph)
and looks like this:


The figure shows a plot of the thrust generated by a particular (variable pitch) propeller as a function of the airspeed (15 through 240 MPH) and power applied to it (250 through 500 HP). The propeller is less efficient at low speed.
Source
-mort

[sendler]

Originally Posted by mort

No.

I worded the question to have an answer of no and that is the perfect chart to show the concept that I am putting forward.
.
But I am still stuck for an easy way to explain it to others. What is the relationship in the physics that explains why the thrust must be less as the speed increases? Which is actually similar to the reduction in rear wheel torque of the bike as speed increases and we are forced to go up through the gears.

[niky]

Originally Posted by mort

For rocketry the specific impulse is often specified. In the wikipedia article you will find this interesting statement:

-mort

This is where the misunderstanding lies. The wikipedia article talks about the velocity of the exhaust in relation to the rocket. Not its velocity in relation to anything else. I'm still waiting for an explanation of what reference you would use to measure exhaust velocity.

There is no air for the rocket to push against in space. At full thrust, at whatever velocity (even zero), the expelled exhaust is moving away from the nozzle faster than whatever exhaust comes after it. So the exhaust can't even push against itself.

Originally Posted by mort

is stupid. Does regenerative braking imply perpetual motion? From looking at the Maxwell's equations basis of the theory it looks like their power efficiency is below 10%
-mort

Regenerative brakes need a medium to push against. And unless you're braking into the solar wind, that won't work (in spaaaaaaace).

[sendler]

Originally Posted by niky

still waiting

Can we come back to this? What do you make of the prop thrust vs speed chart?