General > General Technical Chat
Newton's third law problem.
electrodacus:
--- Quote from: PlainName on November 27, 2022, 07:18:41 pm ---
a) How did you derive those terms? You could post anything and say that's correct, or copy anything and say it's correct "because someone else uses it too", but unless you know how the terms were derived it's meaningless to you.
b) Which term(s) cover the propeller and wheel operation? The propeller, and connection to the wheels, is kind of super-important since it's the thing that makes it work, and yet it doesn't exist in your copied equation. How can that equation possibly account for it, then?
--- End quote ---
a) not only I know how that is derived but also know it is correct as I build things not just play with numbers.
b) The equation for available wind power to any wind power vehicle is independent of the design other than shape and area interacting with air particles.
The wheel propeller connection is a separate issue and using that data you can calculate how much of the available wind power is used to increase vehicle kinetic energy and how much is diverted to propeller witch will use it to store it mainly in the form of pressure differential (this is the explanation for why direct downwind version can exceed wind speed for a limited amount of time).
electrodacus:
--- Quote from: Kleinstein on November 27, 2022, 07:28:23 pm ---The formular for the drag power is the frist one, but this is the power theoretical possible to take from the wind and that is not the power needed for the vehicle to drive. The power for the vehicle is more like the 2nd formular.
It is obvious that the first formula can not be corrent for the power of the vehicle: if the vehicle changes direction from going against the wind to going with the wind it changes from needing power to drive to gaining power from the wind. So the sign in the power has to change at around 0 vehicle speed.
Another way is to look at the force need: mechnical power is force times speed, more or less by definition. The force from the wind is limited and approaching zero speed the power this also has the approach zero.
The relevant question for the vehicle is if the motor can provide sufficient force. The power only determines how fast the vehicle can go. With low speed very little power is sufficient.
--- End quote ---
I do not think that second formula describes anything. I'm fairly certain one or more people that do not understand what power is came up with that as they were likely thinking the correct one outputs a value larger than they expected or wishing.
Yes as you see the sign changes correctly when vehicle speed is zero.
Equation contains this (wind speed-vehicle speed) so max wind power to accelerate the vehicle is proportional with (wind speed - 0)3
If vehicle drives downwind it can be powered by wind at (wind speed - vehicle speed)3
If vehicle wants to drive upwind it requires a power proportional with (wind speed - (-vehicle speed))3 same as saying (wind speed + vehicle speed)3
So is clear from this that no vehicle can exceed wind speed powered only by the wind and can not drive at any speed upwind.
And yes all this is correct. Any wind powered vehicle that exceeds wind speed will do so with either stored energy or another energy source other than the wind and same is true for a vehicle driving upwind at any speed.
When you think at a zero speed vehicle you are wrongly thinking about a vehicle anchored to the ground. But that anchoring to ground just means vehicle is now part of earth.
electrodacus:
--- Quote from: bdunham7 on November 27, 2022, 07:29:11 pm ---You've scoured the internet to find 'examples' where someone has made an error somewhere that supports your argument, but those errors are painfully obvious with just a little bit of a fair-minded consideration.
In the Wolfram example, they clearly have only considered the case where the vehicle speed is the same as the speed of the fluid relative to the body, as in driving in still air. That is obvious simply looking at the equations.
In your other example, if you bother to read their explanations elsewhere on their website under "Vehicle Physics" they say:
Now, the propulsion power can be calculated as a product of driving resistance and vehicle speed:
But from the operation of the malfunctioning calculator app, you can see that they have not incorporated this into their algorithm.
In neither case do they use the equation you claim they do.
The Wolfram example has at least one part right before they go and screw it up. P = FD * Vvehicle
--- End quote ---
It took me all of 3 or 4 minutes to find those two links.
You should read again to understand what that (the thing you highlighted) actually means.
A vehicle on frictionless wheels will be no different from a floating balloon.
Yes when friction is added that is subtracted from the wind power available to vehicle.
They properly incorporated the algorithm and that calculator outputs the correct results.
Look at what is defined as v in that equation as it is not the vehicle speed.
Quote from the text under that equation
"v is speed of the fluid relative to the body"
Which translates for this case in to v = (wind speed - vehicle speed)
bdunham7:
--- Quote from: electrodacus on November 27, 2022, 08:38:18 pm ---You should read again to understand what that (the thing you highlighted) actually means.
A vehicle on frictionless wheels will be no different from a floating balloon.
--- End quote ---
Both of those links purport to calculate the power needed to drive a vehicle against a fluid drag, not the other way around.
electrodacus:
--- Quote from: bdunham7 on November 27, 2022, 09:43:57 pm ---Both of those links purport to calculate the power needed to drive a vehicle against a fluid drag, not the other way around.
--- End quote ---
One link shows the equation the other is a calculator using that equation and that equation will provide you with any question you may have about power needed to overcome drag when driving upwind and it can also calculate the max wind power available to vehicle when driving downwind.
You can just use negative sign in that calculator to calculate wind power available to accelerate so tailwind.
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version