Author Topic: Mess with your minds: A wind powered craft going faster than a tail wind speed.  (Read 147407 times)

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Online fourfathom

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How many here would like to see that happen?

I'd like a ticket for this, please. 

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.
We'll search out every place a sick, twisted, solitary misfit might run to! -- I'll start with Radio Shack.
 

Offline electrodacus

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No vehicle without an external energy source or energy storage device can exceed wind speed direct down wind.

Then your argument has nothing to do with blackbird, you're trying to argue against the accepted scientific facts about sailing boats.

The fact that the boat achieves sailing directly downwind faster than the wind from point A to B through a series of indirect zig-zags is irrelevant, since the length of each zig/zag has no effect on mechanism causing faster than wind travel. 
If the sail boat could change direction in an instant with no time needed to stop, rotate and move the sails, then it could move downwind faster than the wind through a series of zig-zags that were so tiny that they were imperceptible and it would, in effect, be a straight line downwind faster than the wind.

Notice the very important direct downwind or direct upwind in my statement.
While you zig zag a sailboat you can charge the kinetic energy of the boat at higher levels and then you can turn straight and continue for some limited amount of time determined by friction losses and amount of kinetic energy you acquired during travel at an angle to the wind direction.

It is just incorrect to think propeller blades can use wind because they are at an angle as the analogy with a sail traveling at an angle does not exist.
It is a problem of geometry. If you are traveling at 45 degree to the wind direction then wind can still get to your sail as while your speed is above wind speed your speed relative to the wind is not as you take the long route while the wind keeps the direct route.
In case of propeller on a vehicle traveling directly downwind since propeller is fixed to the vehicle no wind can catch any of the blades when vehicle is at wind speed and above.

Offline IanB

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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.
 

Offline electrodacus

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That is the power theoretical available to a wind turbine on the vehicle, but not the power needed to more the vehicle. So a "correct" formular. but for a different problem.
We are discussing the best case here and ignoring frictional losses. And so even in this ideal case no more wind power is available to vehicle than that is calculated with that formula.


0.5 * 1.2 * 0.5 * (64)^3 = 78.6kW of drag with just 0.3kW on input power.
How much more extreme you need the values to be in order to seems shocking.
I am quite shocked by this result - not by the number, but by the units:  the drag is a force and not a power ! :horse:
[/quote]

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.
The results are correct and they should be shocking as you need 78.6kW to drive at 230km/h with no wind and a 0.5m^2 equivalent frontal area
Same will be needed if you only want to move with 1km/h upwind and with a wind speed of 229km/h. There is no difference in power needed if air moves or you move through the air.
Every time wind speed doubles the power needed to overcome drag increases 8x
Why do you think a bike rider capable of 300W can only drive with no wind at 36km/h or so ?  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. Bicycles are mechanically super efficient as much as 95% so the limitation is mostly the air.

Offline electrodacus

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Ok, give me a vote to go ahead and flip 'electrodacus' head this weekend making him convince himself that a wind powered craft going faster than a tail wind speed with his own logic...

How many here would like to see that happen?

I will be very grateful to you if you can manage that as I will learn something that I got wrong.  Unfortunately for you chances are very slim you will be able to do that.
Keep in mind what we are discussing here is blackbird that claims to travel directly downwind faster than wind and so there is no zigzaging to be able to use kinetic energy as energy storage device.
I never claimed Blackbird can not exceed wind speed directly downwind as it clearly can in real world tests including the treadmill model.  My claim is that it can only do that because it can store energy while below wind speed and then that stored energy is what is used to exceed wind speed for a limited amount of time. All my equations perfectly explain what is seen in real world tests.
The formula Derek used is not even close in explaining the real world results as it will predict ever increase acceleration rate rather than the observed slowing down of acceleration rate as you will expect from a energy storage device that it is being discharged. 

Offline electrodacus

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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.

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.

Online bdunham7

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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.

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. 
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline electrodacus

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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.

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

Offline Kleinstein

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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.
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.
 

Offline electrodacus

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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.

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

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).

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.

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.

Online bdunham7

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So this boat with a wind turbine and an energy storage device can finish the race much faster than 1h

Oh my!  What a load of.....????

What was said was that a sailboat without any energy storage can sail the 20 miles in less than an hour.  In fact, give enough distance and a steady 20mph wind from the east, it could sail west indefinitely at more than 20mph.
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline electrodacus

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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 ?

That force will multiply with air speed to get the power available. If you transfer all that force to ground through a bake then vehicle is part of the earth.
As soon as you want the vehicle to move relative to earth vehicle is on his own and it needs that specific power to be able to move against the wind direction.
I'm not going to provide the same examples again but look at the posts I made yesterday to find examples.

Answer this for an ideal case no friction.
How long will it take for a 100kg vehicle to be accelerated from 0m/s to 5m/s with vehicle powered by wind directly downwind and the sail of the vehicle is 1m^2  Wind speed is 10m/s and air density 1.2kg/m^3
Use whatever equations you want answer will be in seconds.


Offline electrodacus

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Oh my!  What a load of.....????

What was said was that a sailboat without any energy storage can sail the 20 miles in less than an hour.  In fact, give enough distance and a steady 20mph wind from the east, it could sail west indefinitely at more than 20mph.

I do not get your reaction. I proved to you that a vehicle that has an energy storage device can get there in less than an hour.
Blackbird has an energy storage device (pressure differential created by the propeller).

Offline Kleinstein

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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.
"Knowing" (or better some intuition) what to expect as the result, however does not mean to put you intuition for the result over the calculation. A carefully done calculation is way more reliable than the intuition. There are some effects in physics that confuse a simple mind and some effects even surprise an expert (though less often). The vehicle in the video is such a case that may cause irritation, though this still a relatively simple one.

Of cause there is not problem using forces and it is the easier way here. Trying to use only power and avoid to use forces at all costs can be quite error prone. It can work in some cases, but gets really tricky with elements like a sail or prop that have less than 100% efficiency.
Of cause one needs to use the right speed and here there seems so be a controversy between electrodacus and essentially the rest of the world, on which speed to use.


That force will multiply with air speed to get the power available. If you transfer all that force to ground through a bake then vehicle is part of the earth.
As soon as you want the vehicle to move relative to earth vehicle is on his own and it needs that specific power to be able to move against the wind direction.
The power actually transfered to the vehincle or needed to more the vehicle is not the same as the power theoretical available from the wind. For the power transfered to the vehicle or needed to move it against the wind, the relevant speed is the speed of the vehicle relative to the ground (the reference system you transfer the force to or calculate the kinetic energy in).
Using the wind speed would need the addition seprate treatment of moving with or against the wind or standing still. Equations in physics are rarely so unsteady, especially not basic mechanics. A high power already at near zero speed would also mean diverging (near infinite) acceleration to store all that power in kinetic energy. So using the full wind speed for the power transferred to the vehicle is obviously wrong.
 

Offline IanB

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I do not get your reaction. I proved to you that a vehicle that has an energy storage device can get there in less than an hour.
Blackbird has an energy storage device (pressure differential created by the propeller).

That's irrelevant. A sailboat, with no turbine, and no energy storage, can sail 20 miles downwind in less than an hour with a 20 mph following wind. It could do this indefinitely.
 

Offline electrodacus

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So using the full wind speed for the power transferred to the vehicle is obviously wrong.

I do not use the full wind power since I use (wind speed - vehicle speed) for a direct downwind vehicle unlike the wrong (vehicle speed - wind speed) Derek used in his video.

You just seems to not understand what air is. Like in my small house at this moment I have over 200kg of air.

Offline electrodacus

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That's irrelevant. A sailboat, with no turbine, and no energy storage, can sail 20 miles downwind in less than an hour with a 20 mph following wind. It could do this indefinitely.

If you are referring to zigzag (not direct down wind travel like blackbird) then you can use the kinetic energy as an energy storage device.

Online bdunham7

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If you are referring to zigzag (not direct down wind travel like blackbird) then you can use the kinetic energy as an energy storage device.

How?  The boat could go 20 miles downwind and 5 to the side without changing direction, or it could make one turn halfway through the course.  In that turn, which only takes a few seconds out of that hour, its speed before the turn and after the turn are the same, so what kinetic energy?
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline electrodacus

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If you are referring to zigzag (not direct down wind travel like blackbird) then you can use the kinetic energy as an energy storage device.

How?  The boat could go 20 miles downwind and 5 to the side without changing direction, or it could make one turn halfway through the course.  In that turn, which only takes a few seconds out of that hour, its speed before the turn and after the turn are the same, so what kinetic energy?

That will not be directly downwind it will be at an angle and it will get to another location compared to someone that went direct down wind.
As for kinetic energy the formula is this  0.5 * mass * (vehicle speed)^2
So you can see the relation between speed and stored kinetic energy
Kinetic energy will not be helpful for a sail type vehicle driving directly down wind but as soon as it drives in other direction than down wind the kinetic energy storage can be used.

Offline Kleinstein

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So using the full wind speed for the power transferred to the vehicle is obviously wrong.
I do not use the full wind power since I use (wind speed - vehicle speed) for a direct downwind vehicle unlike the wrong (vehicle speed - wind speed) Derek used in his video.
At a vehicle speed much smaller than the air speed it does not matter if one subtracts the vehicle speed. At zero vehicle speed your (wrong) formular gives the full power theoretical power in the wind, and this is not the power actually captured. A sail is not at all 100% efficient in converting the energy ! Going against the wind actually needs extra power, while the could provide power if used in a different way. At zero speed the efficency is zero.


When the vehicle moves at the speed of the wind it does not matter if w-v or v-w is used, both would be zero. So even if the sign is wrong it would not make a difference at that point.  When using the correct formular for the power, there is (w-v)² * v and in the square the sign makes no difference.
 

Online bdunham7

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That will not be directly downwind it will be at an angle and it will get to another location compared to someone that went direct down wind.

Then how about the case that I stated where the boat makes one turn at the halfway point and arrives exactly 20 miles downwind in less than an hour? 
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 

Offline electrodacus

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At a vehicle speed much smaller than the air speed it does not matter if one subtracts the vehicle speed. At zero vehicle speed your (wrong) formular gives the full power theoretical power in the wind, and this is not the power actually captured. A sail is not at all 100% efficient in converting the energy ! Going against the wind actually needs extra power, while the could provide power if used in a different way. At zero speed the efficency is zero.


When the vehicle moves at the speed of the wind it does not matter if w-v or v-w is used, both would be zero. So even if the sign is wrong it would not make a difference at that point.  When using the correct formular for the power, there is (w-v)² * v and in the square the sign makes no difference.

That is potential power if you want so it will not do any work but as soon as vehicle moves it will do work.
Obviously at vehicle speed zero w-v or v-w will make no difference other than the sign of potential power showing direction in witch the power can be used if you start moving.

The sign is important as it will show if vehicle accelerates or decelerates and w-v is correct as it will show vehicle can accelerate when vehicle speed is below wind speed directly downwind while using v-w will mean vehicle decelerates while vehicle speed is below wind speed and that will not match any real experiment.
The reason Derek decided to use v-w was to show there is available wind power when vehicle is above wind speed but he needed to use a wrong formula to match his wrong understanding of how the vehicle actually works. Vehicle is not powered by wind directly when vehicle above wind speed (that will be impossible) but it is powered by stored energy.
While that energy that was stored is still wind energy since it is stored energy it will get used up by all the vehicle losses so any real vehicle will slow down after that is used up.

Vehicle is in touch with only two mediums and those have a relative speed of wind speed - ground speed so basically wind speed since that is referenced to ground.
Vehicle weight 100kg vehicle sail area 1m^2 only allowed to drive directly downwind
Relative to ground 0m/s
Wind speed 10m/s
Vehicle speed 0m/s

Potential wind energy in this above mentioned conditions will be
0.5 * 100kg * (10)^2 = 5000Ws = 1.39Wh and this is the max kinetic energy an ideal vehicle can get to since potential wind energy will decrease and vehicle kinetic energy will increase.
 
How come nobody answered the time it will take such a vehicle to get to half the wind speed so 5m/s ?
To be able to correctly answer this question you will need to understand the relation of vehicle speed to available wind power.

I'm sort of an expert in this because I work (my hobby also) in renewable energy storage so I investigated all types of energy storage available and also designed my own wind turbine many years ago. I ended up not using the wind turbine as solar PV made way more economic sense even with relatively good wind resources at my location.
I have designed my own net zero energy house and so both electricity for appliances and heating are supplied by PV solar with energy storage in LiFePO4 and thermal storage in thermal mass (LiFePO4 cost amortisation is around 20cent/kWh while thermal storage is just 1cent/kWh).
I'm not an expert in all area of physics but I have large amounts of experience in energy storage of any type and renewable energy generation.

Offline Kleinstein

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That is potential power if you want so it will not do any work but as soon as vehicle moves it will do work.
Some ( a small fraction when at low speed) of the power would be used if going with the wind, none of this power would be used when moving against the wind. It is wrong to assume that whole available power would be transferred to the vehicle when moving with the wind - only a small fration ( vehicle speed / wind speed) is.

Obviously at vehicle speed zero w-v or v-w will make no difference other than the sign of potential power showing direction in witch the power can be used if you start moving.
How do you define a direction for the power ? power is a scalar value, not a vector so it has no direction. The sign shows wether a system takes up power or gives up power, but this not a direction of geometric sense. A direction in space is a feature of the force.

The sign is important as it will show if vehicle accelerates or decelerates and w-v is correct as it will show vehicle can accelerate when vehicle speed is below wind speed directly downwind while using v-w will mean vehicle decelerates while vehicle speed is below wind speed and that will not match any real experiment.
The importance of the sign is correct, but with the (w-v)³ formular this gets you the wrong result (allways positve) near zero. This is because the fomular is for the theoretical possible power, but not the actual transfered power or power needed to push the vehicle. The formula for the actual power under the condition ( for W>V) , goes with (w-v)²*v gives the right change in sign at aroung v=0 and thus include the transition between against and with the wind.  This formular was never meant to also include the case v > w.  To also incluse the reversal of the wind direction at the vehicle it needs an extra factor for the direction. So the form for the full range is more like (w-v)2 * v * (w-v) / |w-v| for the direction of the drag force.

Anyway for the calculation of the vehicle with the prop we do not really care about a vehicle with only a sail. For the ideal case with a 100% efficient prop the formular for the wind power would come up, as the inverse as the power at least needed for the prop to drive the vehicle.  A real prop would need more power, but can still be good enough.

The case of the backbird vehicle is different from the case of driving against the wind with the wheels driven by the prop / fan. An important point is that the prop is driving the vehicle forward, while the wheels provide the power. One may get a wrong idea here when using the analog to the sailboats. So I consider the sail-boat part rather confusing for understanding the version with the prop.
As the prop is used for propulsion and not to generate the power directly, we actually do not care about how much power the prop in theory could produce. For the way that I find easiest to understand I don't need that formula and I also don't need complications with storred energy.  There are ways to make it work with storred energy, but there are also ways to make it work without.  The picture is relatively simple and does not need a prop with ideal or even good efficiency. I have shown this before, so don't repeat here. The main point that electrodacus tends to get wrong is that the bances of forces decides which way the vehicle can accelerate. There is no such thing as a balance of driving and breaking power - especially not if the efficiency is not 100%.
 

Online bdunham7

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How come nobody answered the time it will take such a vehicle to get to half the wind speed so 5m/s ?

Because it depends on the mass of the vehicle, which can be made arbitrarily small for theoretical calculations.  Some aerogel and carbon fiber if you like.  It doesn't affect the end result, only how long it takes to get there.

Quote
I'm not an expert in all area of physics but I have large amounts of experience in energy storage of any type and renewable energy generation.

Let me guess--you're a completely self-taught iconoclast.  Your constant muddling of terms and concepts is a dead giveaway.  The very basic error in your understanding of mechanics is actually not an uncommon one for beginning physics students, however your intransigence and resistance to the methods and examples typically used to overcome those issues is truly astounding.  I suspect you believe that conventional views on physics are somehow 'wrong' and you have your own personal way of thinking about it that is 'right'.  That would be OK, sometimes I think academics and others can be rigid and pedantic about certain issues that don't matter in nature, but when you come up with a clearly wrong answer, verifiable through experiment, it is time to reexamine your views and methods.  You've actually made some very good models and drawings that quite nicely disprove your points, but then you post them and claim just the opposite.

So here are some things you've gotten very wrong.

1) Conservation of energy by itself is not a law.  You need a lot more definitions and conditions--which are not met in most of your examples and ruminations--before you can use it.

2) 'Stick-slip hysteresis' and all other strange phenomena you refer to are probably real physical effects (if we could figure out exactly what you mean) but can be made arbitrarily small in a theoretical model--and if you think they must be included, you need to quantify them.

3) A sailboat cannot use 'stored kinetic energy' to travel faster than the wind for 45 minutes.  A large oil tanker would have trouble with that, a sailboat will coast to a stop within a few hundred yards (with a buttered teflon hull) and within a few inches (on anything I've sailed on).

4.  The Blackbird can indeed travel straight downwind indefinitely at faster than wind speed. 

Edit:  If anyone thinks they can convince Electrodacus to admit his errors, read this and the subsequent replies first....good luck!

https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3665644/#msg3665644



« Last Edit: December 16, 2021, 10:14:28 pm by bdunham7 »
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 
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Offline thm_w

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People on a motorcycle are shielded by the motorcycle body there is no 0.5m^2 of human exposed to those speeds.

What? Even skydiving where your whole body is exposed does not need shielding at 240km/h, just goggles:

"A stable, freefly, head down position has a terminal speed of around 240-290 km/h (around 150-180 mph). Further minimizing body drag and streamlining the body position allows the skydiver to reach higher speeds in the vicinity of 480 km/h (300 mph)."


The resistance drop on under inflated tires tends towards an exponential curve, not linear and it also depends on the weight of the driver.  I used to be 250lb, an 80psi tire was mushed close to the rip and believe me, I was adding a good additional 50-75 watts just to maintain 25km/h compared to 120psi tires at that weight.

It may have felt like 75W, but even a really poor performing tire at 40psi is sub 30W total: https://www.bicyclerollingresistance.com/cx-gravel-reviews/challenge-grifo-pro
If its near the rim we are probably talking ~15psi, severely under-inflated.
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