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Mess with your minds: A wind powered craft going faster than a tail wind speed.

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electrodacus:

--- Quote from: fourfathom on December 16, 2021, 05:05:32 pm ---Are you building something?  Just don't use a rubber-band instead of gears or chain-drive.  We wouldn't want any of that "stick-slip hysteresis energy storage" to cloud the discussion.

--- End quote ---

I have bad news for you. Both gears and chain-drive have energy storage and stick slip hysteresis. I already demonstrated that in my videos.
For gears only vehicle the energy storage was gravitational as the gear was lifted to charge and fall to discharge and the interaction between gears is documented at 120FPS and is clearly showed during discharge the gear changes the side of the tooth that it engages.

All wheels only vehicle represent the equivalent of upwind blackbird and that one is not using a large storage capacity pressure differential but a very small internal storage in the mechanism either elastic or gravitational that is being charged and discharged multiple times per second and it is smooth out by the kinetic energy once the speed is increased so it will be harder to observe.

Here is just normal zoom out video https://odysee.com/@dacustemp:8/gear-slow30p2:9
The two gear wheels on an axis is the actual vehicle
And here is the zoom out and slowed down footage unfortunately is upside down due to the way it was filmed with my soldering microscope and forgot to flip the image in editing but you can see the effect of discharge at the end of the video when the orange wheel (also the vehicle) changes direction suddenly as the small gear was pushed over the tooth and was falling back down
https://odysee.com/@dacustemp:8/120fps24:9
The teeths are not straight they have an angle so the wheel if locked as it is the case will use the force to climb acquiring gravitational potential energy that is then discharged as it falls back down.

bdunham7:

--- Quote from: electrodacus on December 16, 2021, 05:33:44 pm ---Feel free to use force if that is your favorite but that is how mistakes are made. I personally prefer to use power as it already includes the speed also and there can be no confusion.

--- End quote ---

That's like saying you personally prefer to use watts rather than volts because watts include amps....

The most basic physics instruction will teach you that you need to use the correct units to get the correct answer.  Using the complete form of the units all the way through the equation is how you determine whether 'the units come out'.  Having the units come out to a nonsensical expression indicates that you made an error somewhere, either small typo-like mistake or a fundamental misunderstanding.  Try it--use the complete form of all the units in your equations and at the end you will have some nonsensical result like square watts per meter-second or something like that.

The force of the wind on a sail is expressed in Newtons or the equivalent in another system.  Period.  It doesn't matter whether it is a big sail in a small wind or vice versa.  Now you may protest that a system will behave differently in the two cases as the sail moves, which is true, but that is because the force, expressed only in Newtons, will change differently as the sail and wind speeds change.  You simply cannot express force in energy or power terms as it isn't what the word means--and the units won't come out. 

electrodacus:

--- Quote from: IanB on December 16, 2021, 05:31:39 pm ---We have to put this in terms with no room for misunderstanding.

If the wind is blowing from east to west at 20 mph, then a boat can sail 20 miles due west in less than one hour.

--- End quote ---

That will not mean energy storage was not used.

That 20mph will mean about 8.9m/s
So you can have a wind turbine installed 40% efficient and 1m^2 swept are keep the boat anchored to the ground and wait for say 15 minutes as you still have 45 minutes to get to finish line.
During that time you captured 0.5 * 1.2 * 1 * 8.9^3 * 0.4 * 0.25 = 42.3Wh and you can use that to exceed wind speed so if there is no friction ideal case absolutely no problem to exceed wind speed and maintain that for the remainder of those 45 minutes.
Say boat is 300kg and you need to accelerate that to 40mph (2x the wind speed) 17.8m/s
0.5 * 300 * 17.8^2 = 47526Ws / 3600 = 13.2Wh so plenty of energy to spare from the 42.3Wh if you have no frictional losses ideal case
So this boat with a wind turbine and an energy storage device can finish the race much faster than 1h

Kleinstein:

--- Quote from: electrodacus on December 16, 2021, 05:33:44 pm ---It is the air resistance and if you could move that you can get to almost any speed on a bicycle.
There is no difference between you moving through air or heaving a similar speed headwind.

--- End quote ---
The air resistance is a force. So the force is the same, however that does not mean that the power is the same.
Is there any problem understanding the very basic equation for mechanical power being force time velocity ?

So the power needed is proportional to the speed you move: against the wind or force you need to provide power, at zero speed there is not power needed and with the force and morement there power provided by the force.  The is no need to treat the 3 cases seprate, and no sudden jump in the power going from 0 speed to snails-speed.

I totally agree that the high performance sail on a zig zag corse and the propeller driven vehicle are working somewhat different. For understanding how the blackbird vehicle works, that sail boat part does not really work. At least the usually calculation is different.

The sail boat is only showing that it is in deed possible to get downwind faster than a ballon - though this needs a rather good boad or a sail on wheels / ice to get low friction. So those sail vehicles show that it is indeed possible to still get energy from the wind. So your assuption that it should not be possible at all is already proven wrong. Anyway this would be only an unproven intuition and no way a valid argument.

electrodacus:

--- Quote from: bdunham7 on December 16, 2021, 06:04:34 pm ---
--- Quote from: electrodacus on December 16, 2021, 05:33:44 pm ---Feel free to use force if that is your favorite but that is how mistakes are made. I personally prefer to use power as it already includes the speed also and there can be no confusion.

--- End quote ---

That's like saying you personally prefer to use watts rather than volts because watts include amps....

--- End quote ---

Yes that is a good way of putting things. If you want to solve a problem of conservation of energy is best to look at power rather than voltage or current.
How do you think so many are fooled by motor connected to a generator (those overunity devices).


--- Quote from: bdunham7 on December 16, 2021, 06:04:34 pm ---The most basic physics instruction will teach you that you need to use the correct units to get the correct answer.  Using the complete form of the units all the way through the equation is how you determine whether 'the units come out'.  Having the units come out to a nonsensical expression indicates that you made an error somewhere, either small typo-like mistake or a fundamental misunderstanding.  Try it--use the complete form of all the units in your equations and at the end you will have some nonsensical result like square watts per meter-second or something like that.

The force of the wind on a sail is expressed in Newtons or the equivalent in another system.  Period.  It doesn't matter whether it is a big sail in a small wind or vice versa.  Now you may protest that a system will behave differently in the two cases as the sail moves, which is true, but that is because the force, expressed only in Newtons, will change differently as the sail and wind speeds change.  You simply cannot express force in energy or power terms as it isn't what the word means--and the units won't come out.

--- End quote ---

This are the correct units for solving this particular problem.
And yes I fully agree that you should know what to expect as a result.
There is no problem in using force instead of power but in that case you need to know what is the correct speed you need to multiply with to get the correct result.

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