Nearly did a spit take while doing my math homework. I know it's a theoretical situation, but I think we all know that the efficiency would be zero after the motor came apart...
WTFNearly did a spit take while doing my math homework. I know it's a theoretical situation, but I think we all know that the efficiency would be zero after the motor came apart...
View attachment 1715103668
Excluding catastrophic failure,
What I want to know is how is "c" determined
I mean say that you have a 10/1engine v1/v2 would be 1 to1/10 or 10; then
E%=100 x 1 -{1/10^c}
If I choose .5 for c, then E= 80%
If I choose 1 for c, then E=90%
If I choose 2 for c, then E=99%
If I choose 3 for c, then E=99.9%
But say you have a 8/1 engine,then v1/v2 would be 1 to 1/8= or 8; then
E% =100x 1-{1/8^c}
If I choose .5 for c, then E=75%
If I choose 1 for c, then E=87.5
if I choose 2 for c, then E= 98.44
If I choose 3 for c, then E=99.8%
So it seems like c would have to be less than 2 for sure,
and possibly less even than 1
but greater than 0
So what is"c"?
Yeah... Probability of engine failure!Isn't that a "probability" equation??
Treblig
LMAOSorry, double post.
Yeah OKPhysics vs. Engineering - theoretical discussions vs. real world
Physics vs. Engineering - theoretical discussions vs. real world