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Electroboom: How Right IS Veritasium?! Don't Electrons Push Each Other??
electrodacus:
--- Quote from: dunkemhigh on July 10, 2022, 10:50:58 pm ---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 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.
--- End quote ---
The balls are 1m apart so if vehicle is longer than 1m there may be one or two balls stuck between the sails but they do nothing as they have the same speed as the sails or maybe fall down when says go under due to gravity.
Sails will require energy from somewhere in order to be able to hit the balls. It will be either from some energy storage device if the vehicle is to accelerate forward or from vehicle kinetic energy but that will slow the vehicle down so such a vehicle can never exceed wind speed.
Naej:
--- Quote from: electrodacus on July 10, 2022, 11:41:14 pm ---
--- 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)
--- End quote ---
Did you read the first part?
It says: Let's say efficiency is defined as force*car speed/mechanical power on input so that your argument works.
So it's not efficiency with respect to wind power.
If you have 1000W in wind energy you can for example leave 700W in the air, use 100W for the drag of the vehicle, 100W to compensate tire friction, 100W for accelerating.
So the propeller gets 300W out of 100W in mechanical input, or 300% efficiency.
Note that I used a definition of efficiency to match your use of the word.
--- Quote from: electrodacus on July 10, 2022, 11:41:14 pm ---
--- 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
--- End quote ---
Wrong it's the same.
Relative wind speed is |wind speed - vehicle speed|.
--- Quote from: electrodacus on July 10, 2022, 11:41:14 pm ---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.
--- End quote ---
Yes kinetic energy is always positive.
--- Quote from: electrodacus on July 10, 2022, 11:41:14 pm ---I guess now but you will also claim this equation only applies to blackbird tho the equation contains absolutely nothing specific to blackbird.
--- End quote ---
Wrong again.
--- Quote from: electrodacus on July 10, 2022, 11:41:14 pm ---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.
--- End quote ---
Of course it's not about drag power. It's the kinetic energy in the wind.
--- Quote from: electrodacus on July 10, 2022, 11:41:14 pm ---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.
--- End quote ---
Maybe you should read what I wrote then??? "And the approximation is good when the relative speed is large."
As I explained before - but you won't listen - in still air, you still have some air flowing through a propeller when it is powered, but it is due to power input.
Now I could find a better approximation but since you don't really read what I write it makes little sense.
So where do you disagree then? Either it's the amount of kinetic energy per kg, or it's the mass flow.
gnuarm:
--- Quote from: electrodacus on July 10, 2022, 06:28:11 pm ---
--- Quote from: gnuarm on July 10, 2022, 06:04:21 pm ---The car is going down wind. The wheels on the car are connected to a belt so the top side moves backwards, opposite the direction of travel of the car.
There are sails attached to the belt. When the car moves forward, the belt moves backwards and the sails with it. Since the gearing is 1:1, the sails actually stand still as the car moves under them.
I only want to consider the sails on the top side, so when not on the top side, a mechanism collapses the sails to no size, so they do not interact with the wind.
When the wind pushes on the sails, it will push on the car as a whole. The car will move forward. The belt is geared to the wheels, so moves backwards wrt the car, but stationary wrt the ground, just like the sail. The wind speed blowing on the sail is always the same, no matter how fast the car moves. So the available power from the wind is always the same according to YOUR formula.
The result is the car will be propelled down wind, at a speed determined only by the various power losses vs. the wind power, otherwise having nothing to do with the wind speed. In other words, the car can move faster than the wind, with no other theoretical limitations than the power in the wind vs. the frictional losses.
If you don't like the gearing that leaves the belt stationary, change it to 4:3 gearing so the belt moves backwards at 3/4 speed of the car moving forward. Then the wind relative speed of the sail does not drop to zero until the car is moving at 4 times the speed of the wind. So at some speed, there will be equilibrium between the wind power and the losses in the car. For that speed to be greater than the wind speed, only requires practical design considerations in limiting the losses.
--- End quote ---
I do not not think you fully understand how the vehicle you describe will work.
Seems like what you just did was replace the axial propeller with a savonius type design
--- End quote ---
We don't need to consider any windmill since that would not apply to a car with sails. You do know a windmill is not a sail, right? This model uses sails, as does your equation that you insist on applying to everything. This is very simple to understand and analyze. That's why it is proposed. So, of course you are going to try to change it entirely.
--- Quote ---Forget about air and wind
--- End quote ---
Nope. I've proposed a simple, clear model of a car in the wind traveling down wind that will clearly be capable of going down wind. Let's stick to that.
--- Quote ---and imagine there is a automatic gun shutting 1.2kg balls leaving exactly 1m distance between them at say 6m/s relative to ground so that means each second your vehicle can be hit by six 1.2kg balls traveling at 6m/s
Now one of this balls hits the sail that you mentioned and so vehicle can gain that kinetic energy.
Say your vehicle was stationary first ball hits one of those sails. What happens to the vehicle and the sail relative to the ground ? What direction do they move relative to ground ?
Since you say vehicle is direct down wind the vehicle will move in the same direction the ball was moving and so will the sail even if you you seome gear ratio so that sail moves more than the vehicle or the other way around both sail and vehicle will need to move in the same direction relative to ground.
Only if you are talking about a direct upwind vehicle you can say the sail will move in the opposite direction to vehicle direction.
--- End quote ---
No, you don't understand. Relative to the car, the sail moves toward the wind. With 1:1 gearing relative to the ground, the sail doesn't actually move at all, with 4:3 gearing it moves relative to the ground at 1/4 the speed of the car.
Stop trying to muddy the waters. This car will clearly move down wind, at speeds greater than the wind, only limited by the frictional losses. You insistence in applying an erroneous calculation to the Blackbird car is fixed by this model where your equation does apply, since the wind is blowing into sails, not the air movement produced by the propeller.
electrodacus:
--- Quote from: Naej on July 11, 2022, 12:19:10 am ---Did you read the first part?
It says: Let's say efficiency is defined as force*car speed/mechanical power on input so that your argument works.
So it's not efficiency with respect to wind power.
If you have 1000W in wind energy you can for example leave 700W in the air, use 100W for the drag of the vehicle, 100W to compensate tire friction, 100W for accelerating.
So the propeller gets 300W out of 100W in mechanical input, or 300% efficiency.
Note that I used a definition of efficiency to match your use of the word.
--- End quote ---
Please define exactly the "force*car speed/mechanical power on input"
What applies that force ?
car speed relative to ground or wind?
and what is exactly mechanical power on input?
There is no such thing as 1000W of wind energy (Watt is unit for power not energy)
If mechanical input to a propeller is 100W the output will be in ideal case 100W and in real world more like 70W for an efficient propeller in air.
You have very significant problems in understanding so I was quite wrong about you.
It makes no sense to replay to the rest of your comments as long as you think efficiency above 100% is possible and you confuse power with energy.
gnuarm:
--- Quote from: dunkemhigh on July 10, 2022, 07:23:54 pm ---Not at all. If the ball hits the sail what is there to stop the vehicle heading in the opposite direction? There are no balls hitting that, so the only force is on the sail.
But also doesn't it depend on the gearing? That is, the sub-question is how far the vehicle moves compared to the sail.
--- End quote ---
The point of this vehicle, is to arrange a sail that does not move as fast as the car, so that the power equation he keeps insisting is the right one, really is the right one. So now, the sail moves to the wind end of the car as the car moves in the direction of the wind. As the sail reaches the back of the car (the windward end) it folds up and no longer catches the wind, but moves back to the front with the belt. Other sails continue to unfold at the front of the car (the leeward end) where they start catching the wind again and move to the back of the car.
To keep people from freaking out about a sail that doesn't move at all, consider the gearing to be 4:3, so as the car moves forward (with the wind) by 4 units, the sail moves backwards (toward the wind) on the car by 3 units, which means it moves relative to the ground forward (with the wind) by 1 unit.
The power equation now provides significant power to the sails when the car is moving the same speed as the wind, since the sail is moving with the wind, at only 1/4 the speed of the wind. When the sail is moving with the wind, at half the speed of the wind, it still receives significant power, and the car is now moving at twice the speed of the wind. QED!
I'm not sure, but I think ED is playing dumb so he doesn't have to deal with the reality of this car which clearly can move down wind faster than the wind. He will insist on changing the case and turn it into something where he can obfuscate and confuse the facts.
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