General > General Technical Chat
Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
PlainName:
--- Quote ---If the vehicle moves in the other direction relative to ground compared to the ball that will be a direct upwind so a different type of vehicle that requires a different discussion.
--- End quote ---
Not at all. The problem with it is it's stationary so the sails aren't moving because they aren't being driven. It's effectively a turbine analogue rather than a propeller analogue.
Naej:
--- Quote from: electrodacus on July 10, 2022, 04:55:00 pm ---
--- Quote from: Naej on July 10, 2022, 08:15:51 am ---I didn't change anything and I don't need to change anything.
--- End quote ---
If you say so. The relative wind speed was just wind speed before. But not relevant anyway, I just need a final equation you think describes what happens in reality.
--- End quote ---
Nah I said wind speed IF the car speed is equal to twice the wind speed.
--- Quote from: electrodacus on July 10, 2022, 04:55:00 pm ---
--- Quote from: Naej on July 10, 2022, 08:15:51 am ---Yes. But the recoil means you accelerate the car and it is larger.
--- End quote ---
So will it be able to "recoil" without energy storage. There is no chance to put more power in to propeller than you take at the wheel.
--- End quote ---
Yes, and yes there is a chance.
--- Quote from: electrodacus on July 10, 2022, 04:55:00 pm ---
--- Quote from: Naej on July 10, 2022, 08:15:51 am ---No. What I think is that you get more force, but you don't see the difference between force and energy so ::)
--- End quote ---
What has force to do with anything ?
--- End quote ---
It's called mechanics. Forces mean there's an acceleration.
--- Quote from: electrodacus on July 10, 2022, 04:55:00 pm --- Looking at force alone is a bad idea and that is what likely got Derek in trouble.
The one that needs to see the difference between force and power in not me.
Take the case of vehicle at 2x the wind speed (say wind speed is 10m/s and vehicle speed 20m/s)
Apply a 10N force at the wheel for 1ms (with the generator) and the power you take out ideal case will be 10N * 20m/s = 200W and say you do so for 1ms just so that we can also talk about energy and you have 200W * 0.001s = 0.2 Joules = 0.2Ws
Now take this energy 0.2Ws as that is all you have and apply it for 1ms to the propeller again ideal propeller 100% efficient at propulsion
What you get in therms of propulsion is exactly what you got out by slowing down the vehicle.
So 200W applied at propeller for 1ms will be 20N * (10m/s-20m/s) = -200W so -0.2Ws back into increasing vehicle kinetic energy the negative sign is your choice you can have that for generated power or for propulsion power is non-relevant is just to show direction is different putting in or taking out from the vehicle stored kinetic energy.
So looking at force it is 2x as large but due to speed being half the amount of thrust power and so the amount of kinetic energy you put back in vehicle is the same as the amount you took out in ideal case. In real case it will of course be way less.
--- End quote ---
Let's say efficiency is defined as force*car speed/mechanical power on input so that your argument works.
Now: what if efficiency is 200%?
Of course it's impossible if there's no wind, but there is wind.
If you want to compute the efficiency of a propeller I gave you the formulae.
--- Quote from: electrodacus on July 10, 2022, 04:55:00 pm ---
--- Quote from: Naej on July 10, 2022, 08:15:51 am ---First, it wasn't changed. Second, the formula about kinetic energy was proven a long time ago.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.
--- End quote ---
Please just confirm that the one I wrote is what you mean. If is not write the one you think is correct. Just one equation do not split it in two parts.
Pw = 0.5 * air density * area * (wind speed)2 * (wind speed - vehicle speed)
--- End quote ---
Assuming the two velocities are in the same direction, you have:
Pw approximately equal to 0.5 * air density * area * (wind speed)2 * |wind speed - vehicle speed|
And the approximation is good when the relative speed is large.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.
electrodacus:
--- Quote from: dunkemhigh on July 10, 2022, 10:06:24 pm ---
--- Quote ---If the vehicle moves in the other direction relative to ground compared to the ball that will be a direct upwind so a different type of vehicle that requires a different discussion.
--- End quote ---
Not at all. The problem with it is it's stationary so the sails aren't moving because they aren't being driven. It's effectively a turbine analogue rather than a propeller analogue.
--- End quote ---
Not sure I get what you want to say.
Is a direct upwind a different vehicle than direct downwind ?
If yes then it is important to specify what we are discussing else we will talk about completely different things.
It is also important to use the same reference like vehicle and sail relative to ground not vehicle relative to ground and then sail relative to vehicle.
The sail/sails are part of the vehicle so they will move around the vehicle but together with the vehicle.
So for a direct down wind the 1.2kg ball is hitting the sail transferring the kinetic energy to the vehicle.
When vehicle gets at exactly wind speed (balls speed) there is no way for the ball to heat any part of the vehicle including the moving sails as sails move around the vehicle.
When vehicle moves faster than wind the sails (part of the vehicle) will heat the balls so bals get kinetic energy from the vehicle slowing the vehicle down.
The balls are not a gas so you do not have the advantage of pressure differential.
The example with bals shows why vehicle can not exceed wind speed unless you have stored energy.
PlainName:
--- Quote ---When vehicle gets at exactly wind speed (balls speed) there is no way for the ball to heat any part of the vehicle including the moving sails as sails move around the vehicle.
--- End quote ---
You've got yourself in a mess there. The vehicle is moving at ball speed, right? So the balls are stationary with respect to the vehicle (or vice versa).
Now, the sails are moving relative to the vehicle, so they must be moving relative to the balls as well. How about that, eh!
Nevertheless, somehow you think the balls ain't gonna hit them sails. Well, you're right because it's the sails hitting the balls, but that's just a reference preference. I wonder how you can explain that one away.
electrodacus:
--- Quote from: Naej on July 10, 2022, 10:34:48 pm ---Let's say efficiency is defined as force*car speed/mechanical power on input so that your argument works.
Now: what if efficiency is 200%?
Of course it's impossible if there's no wind, but there is wind.
If you want to compute the efficiency of a propeller I gave you the formulae.
--- End quote ---
??? wow
How will efficiency ever be above 100%
Wind is providing all the input power to vehicle (that if we ignore the pressure differential energy storage as you seems to want to).
So if at any point in time available wind power is say 1000W (just a round number)
You have the choice to
a) use all of it to accelerate the vehicle
b) as it is the case with direct downwind blackbird split it into multiple parts with total still 1000W
An example of splitting the 1000W available will be use 600W to accelerate and take 400W at the wheel to deliver to a 70% efficient propeller so 280W propeller output.
So if the medium is not a compressible fluid the 280W will end up accelerating the vehicle so add to the 600W total 880W worth of acceleration.
880W output/ 1000W input = 88% efficient.
Then there is the real case where those 280W will not all accelerate the vehicle but only small part say 80W with the difference of 200W being used to increase the pressure differential thus stored for later use.
Then you have the same 88% efficiency but only 680W are used for acceleration (vehicle will accelerate slower) but there is that 200W put in to storage than can be used later as it accumulates to exceed wind speed for some limited amount of time.
--- Quote from: Naej on July 10, 2022, 10:34:48 pm ---Assuming the two velocities are in the same direction, you have:
Pw approximately equal to 0.5 * air density * area * (wind speed)2 * |wind speed - vehicle speed|
And the approximation is good when the relative speed is large.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.
--- End quote ---
Another modification but that is fine at least you provided an equation that can be tested.
Will use 10m/s as wind speed
And equivalent area of 1m2
We have vehicle at:
0m/s
Available wind power according to your equation
0.5 * 1.2 * 1* 102 * |10-0| = 600W
According to the equation I claim to be correct
0.5 * 1.2 * 1 * (10-0)3 = 600W
5m/s
your equation
0.5 * 1.2 * 1* 102 * |10-5| = 300W
the one I claim to be correct:
0.5 * 1.2 * 1 * (10-5)3 = 75W
9m/s
your equation
0.5 * 1.2 * 1* 102 * |10-9| = 60W
the one I claim to be correct:
0.5 * 1.2 * 1 * (10-9)3 = 0.6W
10m/s
your equation
0.5 * 1.2 * 1* 102 * |10-10| = 0W
the one I claim to be correct:
0.5 * 1.2 * 1 * (10-10)3 = 0W
15m/s
your equation
0.5 * 1.2 * 1* 102 * |10-15| = 300W
the one I claim to be correct:
0.5 * 1.2 * 1 * (10-15)3 = -75W
20m/s
your equation
0.5 * 1.2 * 1* 102 * |10-20| = 600W
the one I claim to be correct:
0.5 * 1.2 * 1 * (10-20)3 = -600W
30m/s
your equation
0.5 * 1.2 * 1* 102 * |10-30| = 1200W
the one I claim to be correct:
0.5 * 1.2 * 1 * (10-30)3 = -4800W
50m/s
your equation
0.5 * 1.2 * 1* 102 * |10-50| = 2400W
the one I claim to be correct:
0.5 * 1.2 * 1 * (10-50)3 = -38400W
Can you see a problem with results provided by your equation ?
The only two points where the two equations provide the same result is for vehicle speed of 0m/s and for vehicle speed equal with wind speed.
You forced your equation to provide only positive numbers even tho when vehicle speed is above wind speed the direction of air molecules relative to vehicle changes.
I guess now but you will also claim this equation only applies to blackbird tho the equation contains absolutely nothing specific to blackbird.
You will not be able to claim the equation applies to a sail vehicle, to a wind turbine or to a vehicle when we are talking about drag power that is the equation I provided that applies to all that.
Your equation still predicts zero wind power when vehicle speed and wind speed are equal. So how will the vehicle exceed wind speed even the ideal one with no friction ?
What about a real one with say a total of 60W of friction? as according to your equation it can not exceed 9m/s since from there and up to 11m/s wind power available is less than 60W so less than friction losses.
Your equation is not describing what is seen in real world test not the blackbird and not the treadmill model.
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version