I don't understand how this car works.

@DrewDad,

DrewDad wrote:

Well, there's a difference between taking something on faith, which is religion, and actually understanding something, which is science

There is a third option that is neither science nor religion but it just good old common sense in the face of overwhelming evidence ... admitting that it happens but not understanding how. That's what you have finally done on the iceboats and I respect you for that. Kudos.

Now, you say you don't understand where the energy comes from and there's no shame it that either -- but I'm having a hard time giving a crap about spending any more time with you than I already have after your little disrespectful "fly by night" routine over on Physorg. I've been a absolute gentleman here to you and don't deserve that crap at all.

JB

2 Replies

@ThinAirDesigns,

ThinAirDesigns wrote:

but I'm having a hard time giving a crap about spending any more time with you than I already have after your little disrespectful "fly by night" routine over on Physorg. I've been a absolute gentleman here to you and don't deserve that crap at all.

JB

To be fair, that was prior to your involvement in this thread. I apologize for it. I'll return there and apologize as well.

But you are free to involve yourself in the discussion, or not, as you see fit.

1 Reply

@ThinAirDesigns,

ThinAirDesigns wrote:

There is a third option that is neither science nor religion but it just good old common sense in the face of overwhelming evidence ... admitting that it happens but not understanding how.

I'm certainly capable of that, if necessary.

0 Replies

@DrewDad,

DrewDad wrote:

To be fair, that was prior to your involvement in this thread. I apologize for it.

I hadn't realized that the above was true until you just told me (can't tell from the stamps, but I believe you).

Apology accepted (though probably not needed now that I know the timing) and much appreciated.

JB

0 Replies

So, DrewDad - moving forward.

Do you agree that if we were calculating the work that a propeller on a cessna was doing on the air at 5,000ft we would use the distance the propeller travels through the air and not the distance across the ground?

JB

1 Reply

@ThinAirDesigns,

ThinAirDesigns wrote:

I know you don't -- and that's because you are as wrong on the following point as you are on the ice-boats: You claim that the wheels and the prop are calculated over the same distance (Work = Force * Distance). You're just plain wrong.

If you were calculating the power required of the propeller to thrust an airplane downwind you would never use distance over the ground -- you would use distance through the air (what the hell does the ground have to do with a propeller 5 thousand feet in the air?)

Likewise if you calculate the power required of the ddwfttw vehicle's propeller to thrust it downwind you don't use distance over the ground -- you use distance through the air. (what the hell does the ground have to do with a propeller 15ft in the air?)

If you'll give up your hard headed wrongness on the point of the ice-boats and the point of the distance, we'll be done here and you can go on your merry way on the correct side of this brainteaser.

JB

This depends entirely on how you define the system boundaries and exactly what forces you use.

The rate at which work is done by the engine & propeller on the aircraft is = Propeller torque x RPM. This should also equal (assuming incompressible air/i.e. low Mach #) Drag x groundspeed. For any finite interval of time at constant RPM & velocity, simply multiply either expression for power by the time interval.

1 Reply

@georgeob1,

DrewDad wrote:

The rate at which work is done by the engine & propeller on the aircraft is = Propeller torque x RPM.

That's the formula for work done by the engine on the prop -- that's not what we're concerned about for the moment. We're interested in how much work the prop has to do on the air as compared to how much work the wheels do.

Just like at the wheels, it is force*distance, which in prop terms would commonly be called thrust*distance.

You've been arguing that to determine the work done by the prop on the air, in the thrust*distance calcs, one would use the distance the propeller has moved over the ground and I'm trying to figure out why, with the propeller at 5,000ft you would even think to consider distance over the ground and how you think it's relevent.

JB

1 Reply

@ThinAirDesigns,

If you wish to analyze the work done by the airfoils that make up the rotor, then you will have to deal with the forces on the individual blades and the various distances the blade segments travel in their helical path through the air - a rather challenging effort. In any event the result of such a calculation would (except for negligible thermal effects) necessarily equal the integral of torque x RPM over the period in question. In short you are making a distinction without a difference.

The rotor in this case is not powered. Instead it is somewhat like an autogyro of the 1930s which produced lift with an unpowered rotor with the airstream flowing up through the rotor disc.

1 Reply

@georgeob1,

georgeob1 wrote:

If you wish to analyze the work done by the airfoils that make up the rotor, then you will have to deal with the forces on the individual blades and the various distances the blade segments travel in their helical path through the air - a rather challenging effort.

Nope -- thrust*distance will tell you how much work the prop just did on the air. What section of the prop did what is of no importance to us currently -- only what the total output thrust is times the distance covered. Work = force*distance

Quote:

In short you are making a distinction without a difference.

Nonsense -- there are BIG differences. if I have a very inefficient prop, my power calculation of RPM*torque will do little work on the air and much of that power will be wasted. If I have a very efficient prop, I will generate much more thrust from the same RPM and torque.

RPM and torque are a function of propeller design -- for a given amount of thrust I can create a prop with an infinitely variable number of speeds, etc. All unimportant to our current quest.

All we wish to know currently is how much work the prop has to do *on the air*. Thrust*Distance.

Quote:

The rotor in this case is not powered.

Sorry. Wrong. It's a garden variety propeller doing a garden variety thing -- creating thrust from an input torque.

JB

1 Reply

georgeob1 wrote:

The rotor in this case is not powered. Instead it is somewhat like an autogyro of the 1930s which produced lift with an unpowered rotor with the airstream flowing up through the rotor disc.

Just in case you're not up to speed george, contrary to your "unpowered disk" assertion, we're actually applying a torque on the prop shaft greater than almost all performance automobile engines are capable of -- well over 500ft lbs.

http://www.fasterthanthewind.org/2010/05/higher-torque-upgrades.html

1 Reply

@ThinAirDesigns,

The rotor disc of your vehicle has no applied engine torque. Instead it extracts energy from the airstream (via the airfoils )and/or the moving vehicle (via the mechanical coupling with the wheels. The latter, of course represents a loss of energy to the moving vehicle. My analogy is apt.

1 Reply

@georgeob1,

georgeeob1 wrote:

The rotor disc of your vehicle has no applied engine torque. ... My analogy is apt.

ROFLAO!! Yeah, you go ahead and stick with that silliness -- pretending a propeller with >500ftlbs of torque applied to it and turbine with none are 'analagous' in mode.

Carry on.

JB

0 Replies

@ThinAirDesigns,

ThinAirDesigns wrote:

So, DrewDad - moving forward.

Do you agree that if we were calculating the work that a propeller on a cessna was doing on the air at 5,000ft we would use the distance the propeller travels through the air and not the distance across the ground?

JB

Been thinking about this.

There's the work the propeller does on the wind, and there's the work the propeller does on the airplane.

The work the propeller does on the airplane would be measured against ground distance, since the airplane begins and ends stationary on the ground.

To carry the analogy to your craft, there's the work the propeller does on the wind, and there's the work the propeller does on the vehicle. The work the propeller is performing on the craft should be considered in relation to the ground.

2 Replies

@DrewDad,

Quote:

To carry the analogy to your craft, there's the work the propeller does on the wind, and there's the work the propeller does on the vehicle. The work the propeller is performing on the craft should be considered in relation to the ground.

Think this over again carefully DrewDad.

You are suggesting the Physics of the flight of an airplane changes depending on its velocity during take-off... this doesn't make sense.

Aircraft carriers move relative to the ground. The idea that if two identical aircraft--- one that takes off from a moving carrier, and the other that takes off from a runway on the ground... the physics of their flight will be distinguishable in any way once they reach level flight, is problematic.

I would propose that once an aircraft reaches level flight, what the "runway" is doing is completely irrelevant.

For an airplane flying in the air, what matter is what the air is doing (i.e. the velocity of the wind) . (The same goes for the propeller in this example).

0 Replies

@DrewDad,

DrewDad wrote:

There's the work the propeller does on the wind, and there's the work the propeller does on the airplane.

DD - if you wish, ignore the excellent point made above that no one has to ask the pilot where he took off from or where he intends to land before doing the work calcs on the plane -- let me introduce you to a little concept called "equal and opposite".

In steady state flight (which is where we are doing calcs -- pardon me if I didn't make that clear), the propeller can't pull any harder on the air than it pulls on the propeller thrust bearing. If it were pulling harder on one than the other, the plane would be speeding up or slowing down. In the case before us, the work done by the propeller on the air and by the propeller on the plane are equal by definition ... we're steady state.

I must admit, you are one of the more creative fellows when it comes to trying to figure out how to leave the power of the wind out of the equation.

JB

3 Replies

@ThinAirDesigns,

ThinAirDesigns wrote:

the propeller can't pull any harder on the air than it pulls on the propeller thrust bearing. If it were pulling harder on one than the other, the plane would be speeding up or slowing down. In the case before us, the work done by the propeller on the air and by the propeller on the plane are equal by definition ... we're steady state.

OK, you're at steady state.

There is drag on the vehicle due to apparent headwind. There is drag on the vehicle from the wheels (which causes the propeller to spin).

There is propulsion on the vehicle due to the propeller. Some of the propulsion is coming from the wheels forcing the propeller to turn (and generate thrust to the rear). Some of the propulsion is coming from the propeller acting as a sail tacking across the wind.

ThinAirDesigns wrote:

I must admit, you are one of the more creative fellows when it comes to trying to figure out how to leave the power of the wind out of the equation.

I'm trying to figure out how to get the power of the wind into the equation.

1 Reply

@ThinAirDesigns,

If you're trying to say that moving upwind requires more energy than moving downwind, regardless of the type of vehicle you are using, then I will certainly agree with you.

0 Replies

@ThinAirDesigns,

ThinAirDesigns wrote:

georgeob1 wrote:
If you wish to analyze the work done by the airfoils that make up the rotor, then you will have to deal with the forces on the individual blades and the various distances the blade segments travel in their helical path through the air - a rather challenging effort.

Nope -- thrust*distance will tell you how much work the prop just did on the air. What section of the prop did what is of no importance to us currently -- only what the total output thrust is times the distance covered. Work = force*distance

Quote:
In short you are making a distinction without a difference.

Nonsense -- there are BIG differences. if I have a very inefficient prop, my power calculation of RPM*torque will do little work on the air and much of that power will be wasted. If I have a very efficient prop, I will generate much more thrust from the same RPM and torque.

RPM and torque are a function of propeller design -- for a given amount of thrust I can create a prop with an infinitely variable number of speeds, etc. All unimportant to our current quest.

All we wish to know currently is how much work the prop has to do *on the air*. Thrust*Distance.

Quote:
The rotor in this case is not powered.

Sorry. Wrong. It's a garden variety propeller doing a garden variety thing -- creating thrust from an input torque.

JB

You either have a serious problem in communications or seriously deficient in your understanding of both aerodynamics and mechanical systems.

The forces acting on a propeller are transmitted to the aircraft or vehicle in this case by the rotating shaft to which the blades are connected. The propeller can be either driving the airstream (as in a powered rotor) or being driven by it (as in an autogyro or your vehicle). In either case the work done or energy transmitted is the same, whether it is measured at the rotating shaft or at the propeller blades. The simple way to do it is at the shaft, where torque or torsion and RPM can be measured easily and reliably. As a rotating device, every section of the propeller blade travels a different helical distance through the air determined by its radial distance from the hub. Moreover, propeller blades generally have chord, camber and angle of attack that vary significantly with radius - mostly due to the radial variation of speed through the air.

I agree that the propeller on your vehicle can extract energy from the air and transmit it to the vehicle's wheels. With the proper pitch controls and coupling to the shaft driving the wheels, the prop can also store energy in the form of rotational momentum to be used for brief (until the stored energy is exhausted) additional acceleration.

1 Reply

@DrewDad,

DrewDad wrote:

I'm trying to figure out how to get the power of the wind into the equation.

No, you're trying to figure out how to take it OUT because if you do plain jane force*distance calculations on the two mediums - it's all in there already.

You keep making every excuse in the book (including absolutely amazing ones about it mattering where the plane took off and landed) to just not use the simple and proper calcs on the respective mediums. Work = force * distance.

JB

2 Replies

@ThinAirDesigns,

Can you update us on the timeline your group has for releasing official speed numbers, with video?

Would lay this **** to rest, for sure...

Cycloptichorn

1 Reply

I don't understand how this car works.

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