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| Mess with your minds: A wind powered craft going faster than a tail wind speed. |
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| IanB:
--- Quote from: electrodacus on December 20, 2021, 06:37:36 pm ---If you disagree post what you think are the correct ones. I do not make a video without checking multiple times and making sure what I say is correct. I can of course still make mistakes as anyone else. Looking forward to your results for that problem. --- End quote --- No, I will not. Your education is not my problem. You are either trolling, getting people to spend their time posting, to see how long you can make it continue. Or you lack sufficient understanding to support any kind of dialog, and lack any capacity to learn. Either way, you are wasting people's time. |
| electrodacus:
--- Quote from: Kleinstein on December 20, 2021, 09:02:23 pm ---The case with the wheels is a bit different from the prop case, but I think it contains the same difficulty in understanding. As an advantage the basics of mechanics are a bit simpler and less of an approximation. So it may be OK to look again at this simpler picture for the start. To judge if the equations are correct or not the first thing is to make clear what is meant with P_out and P_net mean and the condition for the vehicle. For the condition of the vehicle the situation to look at, the most useful case would be the vehicle standing still in the picture and with F_G = F_M to have a stationary case with no acceleration. So what do P_out and P_net stand for ? --- End quote --- The equations are for the vehicle as the reference so vehicle is considered fixed not moving. Pout is the output on the Motor wheel. Each of the two wheels is on a different independent medium but the wheels can not move relative to each other as they are connected by the body of the vehicle. So taking case A. If you want to take 10W from the front generator wheel then you will end up with a 5N force trying to push vehicle backwards right to left. But you can not have this 5N force unless you also have a 5N equal and opposite force and since this vehicle has just two points of contact the only place that equal and opposite force can be applied is back motor wheel against the Wind treadmill. Pout will for ideal case contain Pin since all power generated to Generator wheel will be delivered with no losses to Motor wheel but to that power there is also the power that Wind treadmill will apply to motor wheel that will be the same force 5N that already exists at M wheel times the speed of the wind treadmill witch is wind speed minus vehicle speed. Since Pin has a different direction than Pout the Pnet shows the net value. If you will want to make calculation of speed maybe plot a speed over time graph then you can select either a speed step or a time step to do the calculation. Say you select a time step of 1ms then you will calculate the kinetic energy of the vehicle after 1ms of 20W net power applied to vehicle and so vehicle kinetic energy will be whatever it was before + 20mWs (20mJ) and then from kinetic energy you can calculate the new speed of the vehicle at t+1ms and use those new values to calculate the new Pnet and then again add that to existing kinetic energy then again calculate the new speed and this way you can plot a speed vs time graph. The main point of this 3 example's was to show that when you exceed wind speed the direction for wind treadmill has changed and thus instead of helping you to accelerate it will help you decelerate. To get around this and ignore the energy storage Derek just decided to reverse the equation and use vehicle speed minus wind speed instead of the correct one. You can not just change the equation when you exceed wind speed and same equation will need to be used for all 3 cases. If Derek was to test his equation for vehicle below wind speed case A he will notice that result will show vehicle deceleration and that will make no sense as that is not what happens in real world. Blackbird is many times pushed started but that is not needed it can start without being pushed. |
| Kleinstein:
What does output of the motor wheel mean ? This is not really clear to me. It that the same as the power needed to supply the motor = the ouput power of the motor ? For now it is just about the equation, so we don' t need to dicuss the 3 cases. We need to understand the equation first. |
| electrodacus:
--- Quote from: Kleinstein on December 20, 2021, 09:59:18 pm ---What does output of the motor wheel mean ? This is not really clear to me. It that the same as the power needed to supply the motor = the ouput power of the motor ? For now it is just about the equation, so we don' t need to dicuss the 3 cases. We need to understand the equation first. --- End quote --- Output at the motor wheel is the input power in the motor from generator wheel plus the power that wind treadmill provide to that same wheel or subtract from the wheel as it is the case at C. There are just two sources of energy available to vehicle one is the road treadmill and one is the wind treadmill. This is because we are looking in the frame of reference of the vehicle so vehicle is the reference thus is not moving. You can salve the same problem in multiple way this is just the one I selected mostly based on the fact that people seems to prefer looking at this problem form the vehicle frame of reference and that is how Derek also solved this same problem. You can look from the ground reference frame where ground / Road treadmill is not moving at 2m/s but vehicle moves forward at 2m/s left to right and in that case the wind treadmill will have moved at 6m/s the wind speed relative to ground/road but it will still be just 4m/s relative to vehicle. The way I like to look at the problem (I'm aware it may look more complicated) was as ground/road being the reference so vehicle was moving at 2m/s meaning vehicle kinetic energy will already be 0.5 * 10kg * (2m/s)^2 = 20Ws Then I can imagine a 10W for 1ms of braking power applied so 10mWs of energy removed from vehicle kinetic energy and stored in some imaginary ideal battery then new vehicle kinetic energy will be 19.99Ws Then if I apply all this 10mWs to the motor wheel the vehicle will gain back the lost kinetic energy 10mWs and be back to where it started 20Ws thus same exact speed but then there is also the wind treadmill that moves at 6m/s relative to ground but just 4m/s relative to vehicle as vehicle already moves in that same direction at 2m/s and so if force on the motor wheel can only be the same 5N there will be an extra 20W for 1ms provided by the wind thus 20mWs so after that 1ms the new kinetic energy for case A will be 20.02Ws so new vehicle speed will be about 2.001m/s |
| Brumby:
You seemed to have "overlooked" this: --- Quote from: Brumby on December 20, 2021, 07:51:33 am --- --- Quote from: electrodacus on December 19, 2021, 05:45:16 am ---Do you understand that power you take from the wheel will brake the vehicle ? --- End quote --- --- Quote from: electrodacus on December 19, 2021, 05:53:12 pm ---If you talk about power and taking a power from the wheel then there is no speed change as power is instantaneous contains no time dimension. --- End quote --- Which is it? You can't have it both ways. --- End quote --- Care to answer? |
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