Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[spork]

I did look up the machine you mentioned, it looks like a really productive unit.

I can't speak to how productive it it - but it's a friggin' blast to drive it up and down steep slopes in the woods!

I'll keep a watch for new and improved progress in the wind machines.

I posted an update to our blog in the last day or two ( www.fasterthanthewind.org ). I've got 4 more days worth of glassing the turbine blades (final coat) before finishing and painting. Then I've got a bit of transmission work to do. As you may have guessed we're not doing anything like the pace we did last year. JB is on the road full-time, and I'm squeezing my efforts in between work and flying. But there's MUCH less to do. I still hope to do some runs in the next several weeks.

 

[RonL]

I can't speak to how productive it it - but it's a friggin' blast to drive it up and down steep slopes in the woods!



I posted an update to our blog in the last day or two ( www.fasterthanthewind.org ). I've got 4 more days worth of glassing the turbine blades (final coat) before finishing and painting. Then I've got a bit of transmission work to do. As you may have guessed we're not doing anything like the pace we did last year. JB is on the road full-time, and I'm squeezing my efforts in between work and flying. But there's MUCH less to do. I still hope to do some runs in the next several weeks.

Hi spork,

http://www.skidsteerhistory.com/How it began.htm

The joy of being paid to work with a machine like the skid loader, is hard to put in words. My first machine was the 1968 M-610 Bobcat, I have witnessed the actual life of these machines and looking at what has developed from those early years until now is hard to believe.
I won't live to see the results of what you guy's are doing, but I think you are the Keller's of wind powered (or assisted) transportation.

Ron

 

[A.T.]

I posted an update to our blog in the last day or two ( www.fasterthanthewind.org ). I've got 4 more days worth of glassing the turbine blades (final coat) before finishing and painting.

Will you post the blade geometry anytime soon?

 

[spork]

Will you post the blade geometry anytime soon?

I'd be happy to. I didn't realize anyone wanted to see it.

c Upwind Car SJM 2X 10 mph wind
vc=(2.0+1)*10*1.467 !vo (f/s)
omega=150 !RPM
rho=.00226 !Density (slug/ft^3)1500 ft MSL
vsnd=1116.5 ! Speed of sound (f/s)
rt=9.25 !Tip radius (ft)
rh=0.5 !Hub radius (ft)
nblade=2 !Blade number
ao=5.73 !Lift curve slope of 2-d section
apzl=-5.0 !zero lift angle of attack (deg)
ld=67 !Section max L/D (6412 @350000 Re)
cl=0.9 !CL at max L/D

! Thrust= 149.975052(lbs) Torque= 351.829559(ft-lb) Power= 10.0482292(HP)
! etat= 0.837301373

rad(ft) chord(ft) Beta(deg) thick(%)
0.937500 1.015620 73.071310 18.490000
1.112500 1.096130 69.482840 17.130000
1.287500 1.176640 66.095340 15.960000
1.462500 1.257150 62.914090 14.940000
1.637500 1.337660 59.938620 14.040000
1.812500 1.418170 57.164120 13.240000
1.987500 1.498680 54.582660 12.530000
2.162500 1.579190 52.184180 12.000000
2.337500 1.662220 49.957450 12.000000
2.512500 1.726720 47.890720 12.000000
2.687500 1.774310 45.972170 12.000000
2.862500 1.806830 44.190270 12.000000
3.037500 1.826130 42.534050 12.000000
3.212500 1.833980 40.993140 12.000000
3.387500 1.832030 39.557910 12.000000
3.562500 1.821790 38.219500 12.000000
3.737500 1.804570 36.969740 12.000000
3.912500 1.781530 35.801190 12.000000
4.087500 1.753660 34.707050 12.000000
4.262500 1.721830 33.681180 12.000000
4.437500 1.686740 32.717960 12.000000
4.612500 1.649010 31.812360 12.000000
4.787500 1.609130 30.959730 12.000000
4.962500 1.567530 30.155920 12.000000
5.137500 1.524530 29.397150 12.000000
5.312500 1.480420 28.679970 12.000000
5.487500 1.435410 28.001270 12.000000
5.662500 1.389660 27.358200 12.000000
5.837500 1.343310 26.748190 12.000000
6.012500 1.296450 26.168870 12.000000
6.187500 1.249120 25.618090 12.000000
6.362500 1.201350 25.093890 12.000000
6.537500 1.153150 24.594480 12.000000
6.712500 1.104470 24.118210 12.000000
6.887500 1.055260 23.663560 12.000000
7.062500 1.005430 23.229150 12.000000
7.237500 0.954870 22.813710 12.000000
7.412500 0.903410 22.416060 12.000000
7.587500 0.850840 22.035120 12.000000
7.762500 0.796880 21.669880 12.000000
7.937500 0.741180 21.319430 12.000000
8.112500 0.683230 20.982910 12.000000
8.287500 0.622320 20.659520 12.000000
8.462500 0.557410 20.348540 12.000000
8.637500 0.486800 20.049270 12.000000
8.812500 0.407420 19.761090 12.000000
8.987500 0.312520 19.483400 12.000000
9.162500 0.178680 19.215650 12.000000

 

[chingel]

I think it is important to understand that it's not just that the propeller is driven by the wheels, it's that if you engage the propeller the wind starts pushing harder on the propeller and the extra wind push drives it forward.

If you engage the propeller at windspeed, it will use as much energy as it puts drag on the wheels, no matter if you multiply the force or not, but the important thing is that the wind will start pushing harder, because it also has to push on the screwing propeller blade.

 

[spork]

I think it is important to understand that it's not just that the propeller is driven by the wheels, it's that if you engage the propeller the wind starts pushing harder on the propeller and the extra wind push drives it forward.

If I were in a motorglider and gliding directly downwind, would you say the tailwind pushes harder on the propeller when the prop is engaged? How about if I were gliding upwind when I engaged the prop?

The motorglider and its prop would see no difference at all (tailwind, headwind, or no wind).

Our cart operates exactly as the motorglider in a tailwind. Th spinning prop interacts with the air it's in in exactly the same way as an airplane prop in flight interacts with the air it's in.

 

[A.T.]

...it will use as much energy as it puts drag on the wheels,...

Comparing energy with a force makes no sense.

...no matter if you multiply the force or not,...

It does matter. Without multiplying the force, the propeller thrust cannot be greater than the wheel drag.

 

[Llyricist]

I'd be happy to. I didn't realize anyone wanted to see it.

c Upwind Car SJM 2X 10 mph wind
vc=(2.0+1)*10*1.467 !vo (f/s)
omega=150 !RPM
rho=.00226 !Density (slug/ft^3)1500 ft MSL
vsnd=1116.5 ! Speed of sound (f/s)
rt=9.25 !Tip radius (ft)
rh=0.5 !Hub radius (ft)
nblade=2 !Blade number
ao=5.73 !Lift curve slope of 2-d section
apzl=-5.0 !zero lift angle of attack (deg)
ld=67 !Section max L/D (6412 @350000 Re)
cl=0.9 !CL at max L/D

! Thrust= 149.975052(lbs) Torque= 351.829559(ft-lb) Power= 10.0482292(HP)
! etat= 0.837301373

rad(ft) chord(ft) Beta(deg) thick(%)
0.937500 1.015620 73.071310 18.490000
...
9.162500 0.178680 19.215650 12.000000

Is that still a NACA 64XX airfoil?

Nevermind, I see it referenced in there.

 

[A.T.]

I'd be happy to. I didn't realize anyone wanted to see it.

c Upwind Car SJM 2X 10 mph wind

Thanks. I just tested it quickly. And in theory it looks like you could reach 2x with:

transmission eff: 0.93
aero drag coefficient: 0.22
rolling drag coefficient: 0.01
frontal area[m^2]: 1.8
mass[kg]: 295

 

[spork]

Thanks. I just tested it quickly. And in theory it looks like you could reach 2x with:

transmission eff: 0.93
aero drag coefficient: 0.22
rolling drag coefficient: 0.01
frontal area[m^2]: 1.8
mass[kg]: 295

Thanks. That's what I'm hoping for (on the optimistic side). What prop efficiency did you use, and what speed ratio (i.e. gearing)?

 

[A.T.]

Thanks. That's what I'm hoping for (on the optimistic side). What prop efficiency did you use, and what speed ratio (i.e. gearing)?

I used a polar that Llyricist created for NACA6412 @450000 Re, so it not quite accurate. I took the gearing that gives 150RPM at 20mph. The inverted prop efficiency at 2xWS given by JavaProp was 81%.

Just a quick test, I will play around more when i have time.

 

[Llyricist]

I used a polar that Llyricist created for NACA6412 @450000 Re, so it not quite accurate. I took the gearing that gives 150RPM at 20mph. The inverted prop efficiency at 2xWS given by JavaProp was 81%.

Just a quick test, I will play around more when i have time.

If you go here:

https://skydrive.live.com/redir.aspx?cid=27579d7d745f6b79&resid=27579D7D745F6B79!103&authkey=qc!eaR1FyTA%24

You can get AF_6.xml, which is the 6412 at Re 350,000

Change the number after the underscore to match the next higher that you actually have.

 

[A.T.]

If you go here:

https://skydrive.live.com/redir.aspx?cid=27579d7d745f6b79&resid=27579D7D745F6B79!103&authkey=qc!eaR1FyTA%24

You can get AF_6.xml, which is the 6412 at Re 350,000

Thanks. It didn't change the results much though. But lowering the Cd to 0.2 and Crr to 0.006 brings you to 2.17 x windspeed.

 

[rcgldr]

The inverted prop efficiency at 2xWS given by JavaProp was 81%.

Isn't the new "prop" going to be used as a turbine to drive the BB upwind? If so, wouldn't the efficiency be limited by Betz law to about 59%? I'm thinking more like 50% overall.

 

[spork]

Isn't the new "prop" going to be used as a turbine to drive the BB upwind? If so, wouldn't the efficiency be limited by Betz law to about 59%? I'm thinking more like 50% overall.

The new "prop" is indeed going to be a turbine - but the Betz limit really doesn't pertain to efficiency per-se. Of course efficiency can be defined a million different ways, but for our purpose I think it makes sense to look at how much wind energy is converted to useful work (torque x rotary speed of turbine) divided by the total energy lost by the wind.

What Betz tells us is that a stream-tube of air of the same diameter as the turbine, can only give up 59% of it's energy to the turbine. This is because the capture area is smaller than the diameter of the turbine (i.e. some of the air ends up going around rather than through), and you can't bring the wind to a full-stop at the turbine (so the air in the wake of the turbine still has some energy).

 

[rcgldr]

The new "prop" is indeed going to be a turbine - but the Betz limit really doesn't pertain to efficiency per-se. Of course efficiency can be defined a million different ways, but for our purpose I think it makes sense to look at how much wind energy is converted to useful work (torque x rotary speed of turbine) divided by the total energy lost by the wind.

As mentioned in a previous post, I'm thinking of efficiency in terms of power output versus power input, and power as force times speed. Using the cart as a frame of reference, power input = relative air speed x force, power output = relative ground speed x force. If overall efficiency was 55%, then with a 2:1 air:ground effective gear ratio, then output force would be (.55 x 2 =) 1.1 x input force, so 0.1 x force factor left over to compensate for rolling resistance and aerodynamic drag with the BB cart moving upwind at 1 x wind speed.

I was thinking that Betz law applied even to open wind mill generators, which would be similar to the turbine usage on the BB cart, but perhaps I'm wrong on this.

 

[spork]

I was thinking that Betz law applied even to open wind mill generators, which would be similar to the turbine usage on the BB cart, but perhaps I'm wrong on this.

It does apply to open windmill generators, and does apply to the cart. But it's really not a measurement of turbine "efficiency" - or at least not a relevant measure of efficiency (since I suppose we can define it as a sort of efficiency if we choose to).

Betz law says that we can't get all the energy out of a specific reference volume of air. It doesn't limit how much of the extracted energy can be put to good use. The latter is the measure of efficiency that matters to us.

 

[rcgldr]

Betz law says that we can't get all the energy out of a specific reference volume of air. It doesn't limit how much of the extracted energy can be put to good use.

I'm wondering how relatively large the turbine will need to be in order to achieve 1x or greater upwind speed with the BB cart. The "efficiency" I'm wondering about is the power the turbine applies to the wheels to drive the cart and turbine against the wind, versus the turbine drag force times the relative (wrt cart) head wind speed.

 

[A.T.]

The inverted prop efficiency at 2xWS given by JavaProp was 81%.

Isn't the new "prop" going to be used as a turbine to drive the BB upwind?

Yes, it's the inverse propeller efficiency:

shaft_power / (turbine_drag * airspeed)

If so, wouldn't the efficiency be limited by Betz law to about 59%?

That's the limit for turbine efficiency defined in the in rest frame of the turbine:

shaft_power / total_kinetic_energy_flow_through_disc

I'm thinking more like 50% overall.

Actually the turbine efficiency of this rotor is much lower, around 12%. But this is not the relevant efficiency for an upwind cart. They are not trying to extract as much energy as possible from a certain volume of air, but rather to achieve the maximal speed. They want as much shaft_power as possible with as little turbine_drag as possible, while for a stationary turbine producing power the turbine_drag is irrelevant.

Imagine you could achieve 100% turbine efficiency and stop the air relative to the turbine. For an upwind car that means you would have not only slowed down the air relative to the ground, but accelerated it in the opposite direction, wasting energy on this. So maximal turbine efficiency is not your goal here. The inverse propeller efficiency is a better indicator, but the cart performance also depends on the other efficiency parameters.

 

[rcgldr]

It's the inverse propeller efficiency:
shaft_power / (turbine_drag * airspeed)

versus Betz law:
shaft_power / total_kinetic_energy_flow_through_disc

This is what I was wondering, if the power to the wheels versus the drag x airspeed could be greater than Betz's 59% factor. If the efficiency is 70% or higher, then greater than 1x wind speed upwind should be achievable with the BB cart.

 

[A.T.]

This is what I was wondering, if the power to the wheels versus the drag x airspeed could be greater than Betz's 59% factor.

Yes sure. For a very slow rotating turbine the inverse propeller eff. is mainly a function of the L/D ratio of the blades. But there are practical limits and minimum power requirements that have to be balanced against this.

 

[chingel]

If I were in a motorglider and gliding directly downwind, would you say the tailwind pushes harder on the propeller when the prop is engaged? How about if I were gliding upwind when I engaged the prop?

The motorglider and its prop would see no difference at all (tailwind, headwind, or no wind).

Our cart operates exactly as the motorglider in a tailwind. Th spinning prop interacts with the air it's in in exactly the same way as an airplane prop in flight interacts with the air it's in.

I'm not sure how a motorglider works, how does it make it's propeller spin?

With the cart that has a screw in a wooden block, when you push the wooden block and the screw isn't turning, you only push a certain amount, but if the screw is turning, you also have to push on the turning parts if the screw isn't slipping in the wood, basically you will be pushing harder when you want to keep the block at the same speed. I think it's the same way with the wind cart. The extra energy to go faster than the wind when you engage the propeller at windspeed doesn't come from multiplying the force, it just makes sure the propeller keeps turning the right way, but because the wind starts pushing harder, ie it also has to work on the rotating parts.

 

[A.T.]

I'm not sure how a motorglider works,

Neither does the air, that interacts with its propeller.

The extra energy to go faster than the wind when you engage the propeller at windspeed doesn't come from multiplying the force, it just makes sure the propeller keeps turning the right way, .

The gearing (multiplying the force) is not only making the propeller turn the right way, it makes the entire cart accelerate the right way. What is "extra energy" anyway? And why do you keep taking about engaging the propeller at windspeed? It is engaged all the time.

 

[Redbelly98]

Betz law says that we can't get all the energy out of a specific reference volume of air. It doesn't limit how much of the extracted energy can be put to good use. The latter is the measure of efficiency that matters to us.

I've been following this discussion as much as I can. Put another way:

• A certain volume of air contains an amount of energy x.
• That air loses an amount of energy y in the turbine.
• An amount of energy z gets converted to mechanical energy.
• z ≤ y ≤ x

Betz law says that y/x (or z/x for that matter) can't be more than 59%. However, z/y is the measure of efficiency used here.

 

[A.T.]

I've been following this discussion as much as I can. Put another way:

• A certain volume of air contains an amount of energy x.
• That air loses an amount of energy y in the turbine.
• An amount of energy z gets converted to mechanical energy.
• z ≤ y ≤ x

All those kinetic energies depend on the reference frame. They are some frames where the turbine puts energy into the air. Your z ≤ y ≤ x is valid only in some reference frames.

Betz law says that y/x (or z/x for that matter) can't be more than 59%.

Betz law applies only in the rest frame of the turbine, which is the rest frame of the upwind cart. In that accelerating frame the kinetic energy contained in a volume of air is not constant. So you can stay well below that 59% at a constant turbine efficiency (around 12%), and still get more and more shaft_power, because the air moves faster and faster. However, that increase in turbine power is countered by the increasing power demand at the wheels, so the theoretical net power is approximately constant and depends mainly on the true wind velocity.

 

[spork]

That's the limit for turbine efficiency defined in the in rest frame of the turbine:

shaft_power / total_kinetic_energy_flow_through_disc

I believe the Betz limit doesn't look at the actual energy flow through the disk, but rather the energy that would flow through the disk if that flow were not disturbed by the disk. I think it uses the disk reference area applied to the free-stream. In reality, some of the energy from the reference area ends up flowing around the disk due to the high pressure upstream of the disk.

I'm not sure how a motorglider works, how does it make it's propeller spin?

That's where the motor comes in.

 

[chingel]

The gearing (multiplying the force) is not only making the propeller turn the right way, it makes the entire cart accelerate the right way. What is "extra energy" anyway? And why do you keep taking about engaging the propeller at windspeed? It is engaged all the time.

Yes you are right the gearing also determines the speed at which the propeller and also the cart will go. It has to have an advantage so that the wind pushing on the propeller wouldn't make it go the other way.

With the extra energy I meant the energy to go faster than windspeed once you are already at windspeed. If you just multiply force by levers, there is no extra energy, but the extra energy comes from the wind which I think starts to push more.

I talked about engaging the propeller at windspeed, because I wanted to understand what happens in such a situation. The only difference I think compared to when it is engaged all the time, is that the propeller already has some momentum at windpseed. The cart should still work when you engage the propeller at windspeed, the propeller just needs some time to speed up, maybe making the cart slow down a little until the propeller is up to speed.

 

[Redbelly98]

All those kinetic energies depend on the reference frame.
.
.
.
Betz law applies only in the rest frame of the turbine...

D'oh, of course!

 

[A.T.]

With the extra energy I meant the energy to go faster than windspeed once you are already at windspeed.

That's not "extra energy". Thats the same energy that is used below wind speed.

but the extra energy comes from the wind which I think starts to push more.

There is no "extra energy". All the energy comes from the true wind being reduced.

 

[chingel]

I mean extra energy compared to when the propeller is not spinning. When you engage the propeller, the wind also works on the propeller blade that is turning, acting like a screw in a wooden block.