DDWFTTW - Directly Downwind Faster Than The Wind

[Rick Willoughby]

Originally Posted by InetRoadkill

I notice that you're mixing rolling resistance with rolling power. Slide 1 shows the energy required at this point is 1.75W and that you have a surplus of power available from a battery. Cool.

Slide 2 shows that the power needed to roll the cart is now 3W. So far so good.

In Slide 3 I have ditched the motor and propeller and fitted a generator to my buggy. I am using another vehicle to push my buggy.

Do you agree with my analysis here?

Rick W.

[InetRoadkill]

Your power demands in slide 3 are now 33W.

Windage power; 1N * 1 m/s = 1W
Rolling power; ((2N + 5N) + 1) * 4 m/s = 32W;

Total power = windage + rolling = 32W + 1W = 33W

[Rick Willoughby]

Originally Posted by InetRoadkill

Your power demands in slide 3 are now 33W.

Windage power; 1N * 1 m/s = 1W
Rolling power; ((2N + 5N) + 1) * 4 m/s = 32W;

Total power = windage + rolling = 32W + 1W = 33W

This is good. It is less than I calculated and you are correct.

In Slide 4 I have put back the motor and propeller and left the generator in place and operating.

Do you agree with Slide 4 and if not why not.

Rick W

[InetRoadkill]

The prop is only producing 9W of power. You're still short 24W.

(9N thrust * 1 m/s apparent wind)

[Rick Willoughby]

Attachment 28499

Originally Posted by InetRoadkill

The prop is only producing 9W of power. You're still short 24W.

(9N thrust * 1 m/s apparent wind)

OK. I can see what you are saying. I was thinking slightly differently along these lines. Lets say I have fitted the generator to the vehicle and it has a clutch so it can be taken in and out of operation.

I am accelerating as we agreed in Slide 2 using the battery and at the instant that the vehicle hits 4m/s road speed (1m/s air speed) I engage the clutch and have a sudden 5N increase in drag due to the generator. You are saying it is obvious that the vehicle will not only stop accelerating but will actually slow down because there is not enough power - that ever present conservation of energy.

Can you give me a more detailed explanation of what will happen as I engage the clutch. I would like some rough numbers so I have included the performance curve for the prop in the data sheet attached. The design point was 12.94W at 1m/s and the motor is a small outrunner synchronous motor operating at fixed frequency so rpm is constant with load. What vehicle speed and prop power will it balance at. Or will it just grind to a halt?


Rick W

[InetRoadkill]

Originally Posted by Rick Willoughby

Attachment 28499

OK. I can see what you are saying. I was thinking slightly differently along these lines. Lets say I have fitted the generator to the vehicle and it has a clutch so it can be taken in and out of operation.

I am accelerating as we agreed in Slide 2 using the battery and at the instant that the vehicle hits 4m/s road speed (1m/s air speed) I engage the clutch and have a sudden 5N increase in drag due to the generator. You are saying it is obvious that the vehicle will not only stop accelerating but will actually slow down because there is not enough power - that ever present conservation of energy.

Can you give me a more detailed explanation of what will happen as I engage the clutch. I would like some rough numbers so I have included the performance curve for the prop in the data sheet attached. The design point was 12.94W at 1m/s and the motor is a small outrunner synchronous motor operating at fixed frequency so rpm is constant with load. What vehicle speed and prop power will it balance at. Or will it just grind to a halt?


Rick W

You need 33N of thrust which is not on the chart. We can extrapolate from the parameters given and get an estimate of 55W of input power needed to produce the necessary thrust. What's going to happen is that since the prop is not producing the required thrust due to the input power being insufficient, it's going to surrender its angular kinetic energy to the thrust it is producing and also to drag and back into the wheels. It will wind down until the energy equations balance again. Where that point happens is hard to tell. Given the interdependency between the wheels and the prop with the possibility of the wheels either trying to over or underdrive the prop I can't make a prediction to the final speed. It will be less than the DW speed. From the videos, I'll assume you found a sweet spot where the prop is able to capture enough energy from the wind to cancel its drag thus allowing the cart to move at near the DW speed.

I can tell you from experience that a windmilling prop is rather draggy. The prop may well start acting as a small spinnaker in this situation.

[Rick Willoughby]

Originally Posted by InetRoadkill

You need 33N of thrust which is not on the chart. ...n.

Where the hell does the 33N come from. My vehicle will accelerate like crazy if I apply that much thrust. It only needs 9N to do 4m/s as you agreed in Slide 3. What has changed all of a sudden.

You already agreed that I only needed 9N of thrust to push the vehicle at 4m/s with the generator connected. Then when I add a propeller that is able to provide 9N of thrust you are saying I actually create a brake that is somehow needing an extra 24N of thrust. What has suddenly changed with my prop that it becomes a brake when I engage the generator. Conditions at the prop haven't changed.

Rick W

[InetRoadkill]

We determined that the energy demands were 33 Watts. The air is moving past the prop at 1 m/s apparent. So ...

Power / velocity = thrust

33 / 1 = 33 N thrust.


edit: the reason it's not 9 is because your pushing against a moving airstream.

[Rick Willoughby]

Originally Posted by InetRoadkill

We determined that the energy demands were 33 Watts. The air is moving past the prop at 1 m/s apparent. So ...

Power / velocity = thrust

33 / 1 = 33 N thrust.


edit: the reason it's not 9 is because your pushing against a moving airstream.

Where is all that extra thrust going. I do not want to accelerate I just want to hold 4m/s?? I only need a thrust of 9N not 33N. Unless you can tell me that I can only conclude your power calculation is nonsense.

Pushing against the moving airstream gets us back to the more power required in an aircraft with a tail wind. I do not think this is the case.

If we have extra force than we need the buggy will accelerate so anything more than 9N means we are accelerating whether the prop is acting on a "moving" airstream or whatever. If the forces do not balance the vehicle is accelerating or decelerating. I think this is Newtons first law. You need to tell me what all that extra force is doing. THe vehicle does not need to accelerate just hold steady speed at 4m/s.

If you fly your plane along at 100kts and then suddenly triple the force it will accelerate like crazy. Why is my buggy any different.

Rick W

[InetRoadkill]

Once again, we determined that the power required was 33W from a couple of posts back. Going back to the car pushing at 4 m/s the required thrust is actually 8.25N (not quite 9N) to achieve this (33W / 4m/s). Note that the speed is 4 m/s with the car pushing. Your airspeed is only 1/4 of that (1 m/s) so the thrust must go up accordingly to produce the required power.

[Rick Willoughby]

Originally Posted by InetRoadkill

Once again, we determined that the power required was 33W from a couple of posts back. Going back to the car pushing at 4 m/s the required thrust is actually 8.25N (not quite 9N) to achieve this (33W / 4m/s). Note that the speed is 4 m/s with the car pushing. Your airspeed is only 1/4 of that (1 m/s) so the thrust must go up accordingly to produce the required power.

Just to be clear on this you are saying that the propeller thrust is going to have to be 33N but the vehicle drag is only 8.25N. By my calculation the excess force is therefore 24.75N. Have I got this right?

You are also saying that extra force is doing nothing because the vehicle is in steady state at 4m/s.

Actually it is Newton's Second Law that needs to be considered:
Second law
Observed from an inertial reference frame, the net force on a particle is proportional to the time rate of change of its linear momentum: F = d(mv)/dt. Momentum mv is the product of mass and velocity. Force and momentum are vector quantities and the resultant force is found from all the forces present by vector addition. This law is often stated as, "F = ma: the net force on an object is equal to the mass of the object multiplied by its acceleration."

How do you reconcile Newton's Second Law with what you are saying?

Rick W

[spork]

Rick, I think there's a critical part you're missing. If you simply start with the assumption that the cart cannot possibly go directly downwind faster than the wind (as inet has), you'll find that you can move numbers around (somewhat randomly) until it satisfies you that your conclusion/assumption is correct.

Don't get too wrapped up with doing the calcs correctly. That could lead to the unacceptable conclusion that DDWFTTW is possible. While that agrees with the physical evidence, it does not agree with inet's assumption - and he's a pilot.

[Rick Willoughby]

Originally Posted by spork

Rick, I think there's a critical part you're missing. If you simply start with the assumption that the cart cannot possibly go directly downwind faster than the wind (as inet has), you'll find that you can move numbers around (somewhat randomly) until it satisfies you that your conclusion/assumption is correct.

Don't get too wrapped up with doing the calcs correctly. That could lead to the unacceptable conclusion that DDWFTTW is possible. While that agrees with the physical evidence, it does not agree with inet's assumption - and he's a pilot.

spork
We have to satisfy a few fundamental laws here and I think Inet knows enough to appreciate that.

We cannot have conservation of energy at the expense of Newton's laws. There has to be a consistent solution that satisfies both.

I just have to get Inet to realise that the wind can provide energy to the system as it does with an aeroplane when flying in a tail wind or when it acts on a spinnaker. After all HE IS A PILOT you know.

Rick W

[Rick Willoughby]

Originally Posted by InetRoadkill

We determined that the energy demands were 33 Watts. The air is moving past the prop at 1 m/s apparent. So ...

Power / velocity = thrust

33 / 1 = 33 N thrust.


edit: the reason it's not 9 is because your pushing against a moving airstream.

IRK
I have tried to set down your theory on forces in the attached slide. Have I got it accurately described?

Rick W

[spork]

Originally Posted by Rick Willoughby

I just have to get Inet to realise that the wind can provide energy to the system as it does with an aeroplane when flying in a tail wind or when it acts on a spinnaker.

Unfortunately, my cynicism gauge is pegging. He put me off from the start. You're clearly doing much better with him, but I still think there's a very high probability he'll prove to be fully logic-proof.

After all HE IS A PILOT you know.

Yeah, me too. I've been both an aero engineer and a pilot for decades. That's why I suggested we avoid any technical terms with him.