- Joined
- Jan 17, 2009
- Messages
- 15,468
- Reaction score
- 233
so, for 7°(exact angle) there is just a little thrust loss.. multiplied by 4
is less than 10°
something good for my ears, and my heart
can you write all the math behind??
i have different numbers
OC = 900
CS = 129
cos = 6.97
and then??
eheheh, i forgot all the math i learned..
wiki helps alot, but can't solve everything
Assuming that all of the motors are canted 7 degrees, the inefficiency (eg: loss of thrust) would be .7% (7/10th of a percent). If you are looking at peak thrust, it would be reduced by that much (multiply peak thrust by .9925 for actual peak thrust (forward).
To get the new (actual forward) thrust curve you would have to integrate that across the actual motor thrust curve (probably not necessary, near as I can figure...)
The loss of thrust at 7 degrees is really very low. The *problem* comes if you have a mis fire.
If you have two motors angled at 7 degrees, but opposite each other (so that they are effectively pushing against each other) there is balance and the overall average thrust vector is UP. If, however, one of those motors doesn't light then you have an angled thrust that the model will have to be able to handle. The classic way to do this is to design such that the motor is angled through the models CG (not sure if this is possible in this design, but it may be).
Hope this helps more than confuses
jim