![]() ![]() Yes, you are - and funny enough, it was you who started the discussion wisecracking about drag rising with the square of velocity I'm discounting your interpretation of whatever Sources you're using, because it simply does not match reality. In no universe does it take 8x the power toįly twice as fast. I've done this sort of testing *a lot* (at least a half dozenĭifferent props on this plane, and again on others). Notice the OSD alreadyĭisplays wattage, and mAh/mile. Measure the current (and thus ultimately power draw). Various throttle settings in different directions (to average out any effects of wind) to In this flight I was specifically testing the efficiency of a specific prop so I was flying level at You tell me what the watts are at 45mph and 90ish mph. I'll post 4 screenshots for the first four numbers I quoted, all from the same flightĪnd *you* calculate the wattage for yourself. Than the amp draw, not at least 2X faster as you're asserting. I specifically said on a "4S pack" which means all the currents I quoted are in the 14-16.5V rangeĪnd as you well know the voltage sags when the current draw goes up which means Watts climbs *slower* Where did I suggest I was comparing current from different voltage packs? It's just like saying on 2s I see 10A, on 4s I see 35A, say, so the power is 3.5 fold - assuming the same voltage sag (very unrealistic), the power will be 2 x 3.5 times, that is 7 times However, all the above assumes that your motor and prop will be as efficient at eight times the power, which is not realistic in most cases So, the upshot is, flying twice as fast requires eight times the power, and your flight-time with the same battery is reduced to one eight - but if you are thinking of covering distance, that will "only" shrink to one fourth. In general, power required rises with the cube of the speed. ![]() If we consider flight time, twice the speed means one eight of the time you can fly, as the power you need for twice the speed rises eightfold, as the force will be rising as the square of the speed, but the speed is factored in once more when calculating the power (power is force times speed or work / energy per unit of time.). You will cover the distance in half the time though would be as efficient at the higher speed, not realistic, more below). ![]() Work (just like energy) is force times distance, so you will have 4 times the energy needed to cover the samee distance (electrical energy from your lipo, assuming the power system etc. The drag force goes up with the square of speed, so twice the speed, four times the force. I think you are aware of the subtleties, but not everyone is, so this might be of help: Drag goes up with the square of velocity so the only way to get great time at speed is withĪ very slippery plane. ![]()
0 Comments
Leave a Reply. |
Details
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |