On Jun 20, 12:18*pm, Brian Whatcott wrote:
es330td wrote:
Nicolas Charmont equipped a Cri-Cri homebuilt with two AMT Olympus
turbojets each producing 51 lbf of thrust that were originally
designed for model airplanes and it flies at 130 mph. ...
Say, for example, one took a 250 lbf engine and mounted it on top a
C172 wing. ...how would the performance of this
plane compare to a standard C172 with a 180 hp O-360 piston engine?
Would it get off the ground in a normal distance or do I need a 5000
foot runway? *Would it be underpowered as compared to the O-360 ...
Thanks.

The underlying principle is straight-forward.
Recips are (more or less) constant max HP devices.
Turbojets are (more or less) constant max thrust devices.

Horsepower is a constant times thrust times speed.

Or if you can work in SI units, it's simpler:
using power in watts
* using thrust *= force in newtons
* * * * * (Newton is about the weight of an apple)
using speed *= meters per second,
THEN power = thrust X speed

On the face of it, that leads to a paradoxical result for recips:

At a slow enough speed, the thrust from a given horsepower is sky-high!
(But props stall out at a fair fraction of stall speed, then thrust
drops off at slower speeds.)
At a HIGH enough speed, the thrust from a given horse power is
teensey-weensey.

Which leads to the first deduction: * * recips are best at slow
airspeeds *and jets are best at high airspeeds.

There's a paradoxical result for turbojets too.....
* * For a constant thrust, the higher the speed, the higher the power it
represents!

Now an example - the one you offered:
2 X 51 lbs thrust allows a top speed of 130 MPH from a cri-cri

How many horses does this mean at this particular speed?
(I'll convert to SI for convenience)
51 lb = 51/2.2 kg = 51/2.2 X 9.8 * newtons = 227 newtons.
130 MPH = 130 / (60X60) * *miles/sec =
130 X 1760 X 36/39.37 /(60X60) m/s = 58 meters/sec

Remembering power = thrust X speed and we already
know speed and thrust, so we can find power, like this:
power = 2 X 227 X 58 = 26.4 kW = 26.4 / 0.760 HP = 34.6 HP

Is that ALL? It implies the cri-cri is a lowish drag machine!
But there are 1-seaters that make 190 MPH on a VW of similar power.

Now, we ought to work your other example: the C-172 with a 360 engine.
This is (say) 180 HP. Can't remember the top speed of the C-172,
so I'll guess 135MPH

Here we have numbers for power and speed, so we can work out thrust.

I'll do it again in SI for convenience:
135 MPH = 60.2 m/s
180 HP = 137.3 kW
so THRUST = power / speed = 137300 / 60.2 = 2281 newtons
2281 newtons = 2281/9.8 * X *2.2 lbs =
512 lbsf. * *As much as THAT?
This implies that the C-172 is a pretty draggy airframe,
about four times as draggy as a cri-cri anyway.....

Well, by now, we are about ready to guess an answer to your question:
how fast will a C172 go with HALF the thrust it has at top speed with a
180HP engine?
Almost ready.
We also need to remember that air drag varies as the square of speed.
So we can say that the thrust needed at any speed is the drag at that
speed, *and for the C-172, drag is 512 lb at 135 MPH
so at some speed V, the drag is 512 times *(V/135) squared

For 250 lbf thrust, we get 250 = 512 times (V/135)^2
Rearranging:
(V/135)^2 = 250/512
V/135 = 0.70
so V = 94 MPH top speed.

Not much of a range between stall and top speed in this low power
c-172 huh?
To finish up, we can work out the power delivered to a C-172 by some jet
at 94MPH top speed

power = speed times thrust.
power = 94MPH X 250 LBF X *some unit conversion constant
We know 180HP *= 135MPH X 512 LBF x the same conversion constant

So the conversion constant is 180 / (135X512)

And FINALLY:
* the power needed to drive a C-172 at a top speed of 94 MPH =
speed X thrust X conversion constant =
94 X 250 X 180/(135X512) = 61 HP

This is a third of the horses to get over 2/3 the top speed!

Hope this helps

Brian Whatcott
Altus OK

What a great post. It took me a minute to get back into physics mode
mentally but this was a great explanation.