Timeline for Lift to drag ratio for a given glider
Current License: CC BY-SA 4.0
9 events
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Oct 28 at 15:52 | history | edited | Rob McDonald | CC BY-SA 4.0 |
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Oct 28 at 15:51 | comment | added | Rob McDonald | If you run through the math and the free body diagram, instead of $L=W$, you get $L=W cos(\theta)$. When you carry through to $L/D$, you get a sine over cosine that turns into a tangent and cancels out -- so $L/D$ is the glide slope without approximation. However, when I suggested you use $C_L=W/(q\,S)$ to get to airspeed, there is a small angle approximation embedded there which could be resolved with a little more work. | |
Oct 28 at 15:46 | comment | added | Rob McDonald | @DarthPseudonym If he knew the interview question ahead of time, I would agree. However, since it seems the interview is over, I appreciate someone trying to learn from the experience. | |
Oct 28 at 15:45 | history | edited | Rob McDonald | CC BY-SA 4.0 |
edited body
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Oct 28 at 15:33 | comment | added | Darth Pseudonym | "Please do my job interview for me" is honestly even worse than "please do my homework for me". The whole point of the exercise is to see what you can do, not what the internet can do for you. | |
Oct 28 at 12:12 | comment | added | Green Lantern | Thank you. Actually, I encountered this problem during a job interview. I'm new to this field and platform, so I hope it’s okay to ask – and please excuse my likely basic questions. Could you clarify why the solution didn’t consider the gliding angle? Also, when you mentioned calculating the cruising CL, did you mean for gliding conditions? | |
Oct 28 at 10:18 | comment | added | DeltaLima♦ | Hi Rob, I converted the math to MathJax typesetting. Can you do a quick check to make sure I didn't make any mistakes in the conversion? | |
Oct 28 at 10:17 | history | edited | DeltaLima♦ | CC BY-SA 4.0 |
Converted math to MathJax typesetting
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Oct 27 at 20:56 | history | answered | Rob McDonald | CC BY-SA 4.0 |