Reverse Thrust Calculations.....
Well Chris, your math is pretty darn good if you really don’t have access to trig. I get the same 200 Kts (obviously, from 100 * cos(60) + 150) and 23.4 degrees (vice your 25 degrees) using your method, and it does make some sense to me.
I’ll perhaps add a caveat though. What if we used the same 150 Kts aircraft TAS and 100 Kts exhaust gas speed, but choose not to deploy the thrust reversers? In this case, the gases start out at 0 Kts relative to the Earth, and after exiting the aircraft end up with 50 kts forward speed relative to the Earth. This means the aircraft would slow down without deploying the thrust reversers at all!!
However, perhaps more realistic figures such as 300 Kts gas speed, and abort from 100 kts would mean your calculation does show, based on a my chosen 60 deg off reverse direction, that indeed there would be more reverse thrust at higher speeds than lower speeds. Maybe when I’m more bored I’ll run a spreadsheet on it.
So I understand your comment about the improved direction of the reaction force. I’ll have to think more about your comment about the strength of the reaction force, you said “its strength is much higher (proportional to the square of the gas speed)”. But if we used 100 Kts aircraft speed, the strength would be 150 Kts, so I’m not seeing the squared relationship. Again, maybe my spreadsheet will address this. Let me know if you’d like to see it when/if I complete it.
Thanks for your response, it sent me in the right direction (pun intended). Of course, We’re causing great anguish to SNS3Guppy…….