Quote from
hawk37 (Chris, can you explain a bit further?):
Assume the thrust angle of the exhaust is 60 degrees off of the direction of travel. Then surely the deceleration force is cos(60) * thrust, or .5 * thrust, regardless of the aircraft's forward speed. No?
[Unquote]
Hi, I'll give it a go.
You are spot-on for the case when the aircraft is stationary. As to why it's different with forward speed, I'm going to try and simplify the applied-maths, but can't avoid using the concept of vectors.
Let's stick at first with your figures, for sake of argument. Imagine the aircraft is pointing North (000 degrees), and the wind is calm.
In the stationary case, let us consider a portion of the mass of the reverser
exhaust gas that is coming out horizontally from the R/H side of the thrust-reverser assembly.
The gas is travelling on a
heading and track of 060 at an unknown speed (say, 100 kts), thus creating a
reaction force (vector) in a direction of 240 degrees. This would tend to push the aeroplane backwards (provided the nosewheel steering remained straight).
Now let's have the aeroplane rejecting a take-off at 150 kts TAS and G/S (wind still calm). Compared with the first case - because it is now coming from a moving aircraft - the exhaust air already has a
forward (000 degrees) VECTOR of 150 kts
in relation to the earth before it leaves the reverser assembly, i.e., 000deg/150kts.
The reverser will impart to it the same VECTOR as before (060deg/100kts),
in relation to the aircraft.
If you combine these two vectors, the RESULTANT
in relation to the earth is roughly (in the absence of any trig. tables, I'm afraid) 025deg/200kts.
This produces a
reaction force in a direction of 205 degrees, better than in the stationary case.
In addition to the improved
direction of the reaction force, its
strength is much higher (proportional to the square of the gas speed). As the gas speed is doubled to 200kts - compared with 100 kts - the strength of the deceleration force vector is 4-times as strong as in the stationary case.
And although I asked if we could consider only one portion of the exhaust gas, it is clear that all of it will work in a similar fashion.
I think
SNS3Guppy is going to demolish the figures we are using, however. He will point out that your hypothetical figure of the exhaust gas travelling
"60 degrees off the direction of travel" is quite unrealistic, particularly for a cascade fan-reverser. And the core air is not reversed at all, remember.
But I hope he will acknowledge that a conservative figure of 90 degrees "off the direction of travel", becomes much lower when you apply the forward vector generated by the aircraft's speed. That is why it's so important to select reverse at high speed. I am convinced he is wrong, however, to put all the braking down to "intake drag".
Propellers can generate reverse thrust vastly more efficiently than jets, particularly at low-speed/stationary, as in the Hercules. But the other reason we do not reverse jets off the stand is the problem of dangerous flying debris and ingestion of same into the fan intake.
Hope this helps.