Wind Power Vehicle Traveling Down Wind Faster Than The Wind

[A.T.]

Certain high-performance sailboats can travel faster than the wind when sailing downwind due to the apparent wind created by their speed. These boats don't travel directly downwind, however.

The spinning airfoils don't travel directly downwind either, even if the vehicle does. The aeordynamics are quite similar to a sailcraft.

Sail craft on broad reach:

[PLAIN]http://img253.imageshack.us/img253/6694/downwindvectorsen3.png

Propeller blade:

[PLAIN]http://img811.imageshack.us/img811/4922/propellervectors.png

The transition:

https://www.youtube.com/watch?v=UGRFb8yNtBo

I don't fully believe they get any more speed from the apparent wind, but I did read this once.

Statements like "getting speed from apparent wind" or "making your own wind" are of course folksy talk that doesn't explain anything. What you can say though is:
- The force on the sail is created by the apparent wind.
- The energy comes from reducing the true wind.

 

[chingel]

This could work (depends on how much drag there is between parachute and ground), but the yo-yo eventually reaches the parachute and the process stops.

In case it wasn't made clear, when the cart is moving forwards, the wheels on the cart are turning the propeller against the wind, so that the propeller is generating thrust which pushes the air backwards (or slows it down). The external forces are the air pushing forwards against the propeller, and the ground pushing backwards against the wheels (the wheels are using a braking force to drive the propeller). The effective gearing combined with a tailwind allow the force with the air to be greater, but at a lower speed, than the opposing force with the ground, while still allowing for some losses in the system.

I think that it's not good to think that the wheels are making the propeller thrust, but rather that the propeller is constrained by the wheels to turn if it is moving forward, and the wind pushes it forward by the propeller and since the propeller is constrained to turn if it moves forward, the wind also pushes on the propeller screwing up the wind.

Like with the screw in a jelly or a wooden block example. The wooden block is attached to the screw. If the screw doesn't turn, the block moves the screw and the cart goes at the block speed, but if the screw also screws, the block pushes on the screw to get up to block speed, but it also pushed the screw because it is rotating. Just that the force from the wheels turning the screw out has to be stronger than the force with which the block is trying to screw the screw in by pushing on it.

I realized that with the ruler example, the center of the wheel is actually moving slower than the point where the force is applied by the bigger wheel, because the wheels are turning clockwise and therefore the point where the force is applied is moving at small wheel speed plus the rotating speed, so viewing it as a lever where the ground is the pivot point should still be correct. It is just that the big wheel which applies the force is also moving and rotating at the same time, so the speed it applies is the ruler cart's speed plus the rotating speed of the big wheel.

 

[spork]

I don't fully believe they get any more speed from the apparent wind, but I did read this once. If they could gain speed from their speed, then they would continue to speed up indefinitely, so it's kind of violating the laws more than perpetual motion does.

Like I said, it gets to be an almost philosophical question. You do use the apparent wind to develop more speed, but it can be shown that it's the geometry that matters. You can't just produce apparent wind from your motion and use it to keep you going.

In kitesurfing, the effect of apparent wind is extremely noticeable in a couple of ways. When we stroke the kite through the "wind window" on light wind days, you can feel a tremendous amount of added pull. Also, once you get the board up on a plane and get going on a light wind day, you can feel a lot of pull - that goes away when you slow down. But there really is no free lunch. You can only exploit the energy of the true wind. Interestingly though, it's possible to exploit much more energy from the true wind when you're speeding through it - and thus affecting much more air.

I think that it's not good to think that the wheels are making the propeller thrust, but rather that the propeller is constrained by the wheels to turn if it is moving forward, and the wind pushes it forward by the propeller and since the propeller is constrained to turn if it moves forward, the wind also pushes on the propeller screwing up the wind.

We can think of the wheels as providing that constraint as you say, but it's absolutely key to recognize that the torque that turns the prop shaft absolutely comes from the wheels. It's also misleading to think of the wind as pushing on the back of the prop when the vehicle is going faster than wind speed. The wind pushes on the back of the prop in precisely the same way that wind pushes on the back of an airplane prop that happens to be going downwind (or upwind, or in no wind...).

 

[rcgldr]

The spinning airfoils don't travel directly downwind either, even if the vehicle does.

In the case of a DDWFTTW cart, spinning airfoils are more efficient, but not a requirement. Instead of a propeller, some type of ducted paddle wheel like device could be used, but it wouldn't go as fast since paddle wheels aren't as efficient as propellers, but the operating surfaces of the paddle wheel would be moving directly upwind.

A really crude cart could have the wheels drive a treadmill so that it's upper surface moves upwind, with a couple of parachutes attached. Each parachute would be opened up when moving upwind, and closed when moving downwind. Again it wouldn't be efficient, but it could work. It would be similar to the yo-yo pulled by a parachute example.

Sailcraft.- The force on the sail is created by the apparent wind. The energy comes from reducing the true wind.

The thrust from a sail is due to diversion of the crosswind component of apparent wind. A sail can't generate thrust from the headwind component of apparent wind.

I think that it's not good to think that the wheels are making the propeller thrust, but rather that the propeller is constrained by the wheels to turn if it is moving forward.

The thrust from the propeller is related to it's speed, pitch, and the apparent head wind (so the true wind is a factor). The torque required to drive the propeller is related to the thrust, and that torque has to be supplied by the wheels, with effective gearing used to multiply the torque from the wheels while dividing the speed from the wheels (taking advantage of the fact that from a cart's frame of reference the ground speed is greater than the wind speed (when there is a tail wind)).

I realized that with the ruler example, the center of the wheel is actually moving slower than the point where the force is applied by the bigger wheel, because the wheels are turning clockwise and therefore the point where the force is applied is moving at small wheel speed plus the rotating speed, so viewing it as a lever where the ground is the pivot point should still be correct. It is just that the big wheel which applies the force is also moving and rotating at the same time, so the speed it applies is the ruler cart's speed plus the rotating speed of the big wheel.

I'm not sure at what you're getting at here with the ground at pivot point. The key factor is the ratio of the inner versus outer diameter of the smaller wheels. In the case of the video cart, the ratio is about 1:2, so the cart moves at 2x ruler speed or -1x ground speed (depending on which is moved). If the ratio was 2:3, the cart would move at 3x ruler speed or -2x ground speed.

 

[OmCheeto]

Here some material on this:

# ^ Drela, Mark. "Dead-Downwind Faster Than The Wind (DFTTW) Analysis". http://www.boatdesign.net/forums/attachments/propulsion/28167d1231128492-ddwfttw-directly-downwind-faster-than-wind-ddw2.pdf.

# ^ Gaunaa, Mac; Øye, Stig; Mikkelsen, Robert (2009). "Theory and Design of Flow Driven Vehicles Using Rotors for Energy Conversion". Marseille, France: Proceedings EWEC 2009. http://orbit.dtu.dk/getResource?recordId=241183&objectId=2&versionId=1.

# ^ Bauer, Andrew (1969). "Faster Than The Wind". Marina del Rey, California: First AIAA. Symposium on Sailing. http://projects.m-qp-m.us/donkeypus...aster-Than-The-Wind-The-Ancient-Interface.pdf.

Excellent! Thank you. This should be fun.

One thing I don't remember seeing is an explanation of the offset rear wheels which they claim is to counter the effect of prop torque.

http://www.blueplanettimes.com/wp-content/uploads/2010/08/axle.jpg
"[URL
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[A.T.]

One thing I don't remember seeing is an explanation of the offset rear wheels which they claim is to counter the effect of prop torque.

That is pretty much the explanation. The aerodynamic forces on the propeller create a torque opposite to the propeller rotation. This torque is countered by the uneven vertical load distribution among the back wheels. To prevent that load distribution from getting too unequal, you make the lever arm of one wheel longer.

 

[OmCheeto]

That is pretty much the explanation. The aerodynamic forces on the propeller create a torque opposite to the propeller rotation. This torque is countered by the uneven vertical load distribution among the back wheels. To prevent that load distribution from getting too unequal, you make the lever arm of one wheel longer.

It's difficult for me to visualize how and where this prop torque is transmitted to the vehicle. Is it at the prop hub or at the axle hub? I don't remember studying the physics of chains on chain driven vehicles. I suppose I'll have to back up and figure out how a bicycle works first.

But as I said, I don't have time to work on this except on weekends.

 

[rcgldr]

The torque at the prop and wheels results in a differential in the chain tension, with the the segment of chain going from the front of the wheel sprocket to the right of the prop sprocket having high tension. This results in a differential in compression of the posts that support the prop axis, with the right post having higher compression. The wheel sprocket appears to be offset to the right so that the high tension part of the chain is more vertical, but that doesn't elminate the differential in the post compression.

 

[A.T.]

It's difficult for me to visualize how and where this prop torque is transmitted to the vehicle.

The prop is already a part of the vehicle. The sum of all external torques around the vehicle's longitudinal axis must be zero. If you consider the relevant external force components (projections onto the frontal plane) you will see that the external forces on the two airfoils create an longitudinal torque, and the only thing that can counter it are the normal forces on the wheels.

 

[rcgldr]

The prop is already a part of the vehicle. The sum of all external torques around the vehicle's longitudinal axis must be zero. If you consider the relevant external force components (projections onto the frontal plane) you will see that the external forces on the two airfoils create an longitudinal torque, and the only thing that can counter it are the normal forces on the wheels.

I think he understands that. He's asking how that torque is being transferred to the vehicle.

Looking at the vehicle from the back, the air generates a counter-clockwise external torque onto the propeller, and the normal force from the ground and position of the wheels generates an external clockwise torque equal and opposing to the external counter-clockwise torque that the air exterts onto the propeller.

Internally, the sprocket and chain produce an clockwise torque onto the propeller equal and opposing to the counter-clockwise torque that the air exerts onto the propeller. The differential in tension in the chain produces a braking torque on the wheels (forward force applied to the ground), with an equal and opposing torque due to the external backwards force applied by the ground to the wheels. The two columns supporting the prop axis are also affected by the position and differential in chain tension, as well as the torque related to the normal force at the wheels, but I'm not sure how this translates into lateral loads near the hub of the propeller, or internal torques.

 

[spork]

...but I'm not sure how this translates into lateral loads near the hub of the propeller, or internal torques.

To first order there are no lateral loads at the prop hub. Due to gradient and misalignments, there will certainly be some lateral loading there.

 

[rcgldr]

To first order there are no lateral loads at the prop hub. Due to gradient and misalignments, there will certainly be some lateral loading there.

Tension in the chain should cause a downward load at the prop axis and an upward load at the wheel axis. The supporting colums should experience a compressive load to counter this.

I'm wondering how the clockwise torque related to the normal force from the wheels is transmitted through the vehicle to end up opposing the counter clockwise torque from the air onto the propeller.

 

[shardbearer]

Im not following this lengthy discussion, but I feel I might as well contribute.
I have personal experience with hydrofoil sailboats going up to 1.5 times the wind speed, and I can assure you all that it is definitely possible.

 

[mheslep]

Im not following this lengthy discussion, but I feel I might as well contribute.
I have personal experience with hydrofoil sailboats going up to 1.5 times the wind speed, and I can assure you all that it is definitely possible.

As the title states, the trick here is traveling directly down wind, without tacking, faster than the wind which with a sail alone is not possible in a traditional sail boat on a level surface.

 

[rcgldr]

I can assure you all that it is definitely possible.

Both model and full scale carts have demonstrated faster than wind speed, with the full scale cart getting a certified run of 2.8x wind speed. Most of the discussion in this thread is about how these carts work, the math, physics, design, and the details.

 

[OmCheeto]

Tension in the chain should cause a downward load at the prop axis and an upward load at the wheel axis. The supporting colums should experience a compressive load to counter this.

I'm wondering how the clockwise torque related to the normal force from the wheels is transmitted through the vehicle to end up opposing the counter clockwise torque from the air onto the propeller.

Excellent! Another wonderer.

Why would there be a 60% difference in axle length from centerline if there was not a major torque going on.

As I've stated from almost the beginning of these threads, this is a most peculiar and delightful problem in mechanics, and worthy of study in any university.

WTG San Jose State University!

 

[ThinAirDesign]

Why would there be a 60% difference in axle length from centerline if there was not a major torque going on.

Why would you think there was not a major torque going on when there is a 60% difference in axle length from centerline?

JB

 

[spork]

Why would there be a 60% difference in axle length from centerline if there was not a major torque going on.

There's very definitely plenty of torque going on. Just about enough to turn the cart over before we lengthened the left axle in fact.

WTG San Jose State University!

Much to our dismay - SJSU had very little to do with this project.

 

[A.T.]

I'm wondering how the clockwise torque related to the normal force from the wheels is transmitted through the vehicle to end up opposing the counter clockwise torque from the air onto the propeller.

In simple terms: The chain is trying to turn the wheel axis anti clockwise around the prop axis, while the normal forces from the wheels are in combination trying to turn the wheel axis clockwise around the prop axis.

Let's look from the back and use the prop center as the reference point in our 2D projection for all the torques:

The force in the chain is creating a clockwise torque at the prop axis, but also an anti clockwise on the wheel axis. And by the later I really mean: A torque at the wheel axis in respect to the prop center, parallel to the vehicles longitudinal axis (not to the wheel axis itself). This longitudinal torque at the wheel axis in respect to the prop center is independent of where the wheel axis sprocket is placed, and is countered by the net torque from the wheel normal forces, in respect to the prop center.

 

[OmCheeto]

Much to our dismay - SJSU had very little to do with this project.

? ?

So they dropped out of the project?

http://www.fasterthanthewind.org/2009/10/rick-and-i-had-good-lunch-meeting-with.html"
Rick and I had a good lunch meeting with Professor Mourtos on Tuesday. He introduced us to a late addition to the SJSU design team. Shethal Thomas is a graduate student in the aerospace engineering program at the University and we're excited to have her on board.

After lunch, we dropped by the Aerospace Engineering Lab and located a good spot to use as a workspace for building the propeller. I'll try to get some pictures once we get the space set up.

JB

 

[Subductionzon]

From my understanding the students quickly lost interest in the project. Too bad for them, they could have been part of a truly fascinating build.

 

[rcgldr]

The force in the chain is creating a clockwise torque at the prop axis, but also an anti clockwise on the wheel axis.

I got that, but take the case where the right side of the prop sprocket is directly above the front of wheel axis sprocket, so that the chain under tension takes a vertical run (just twisting 90 degrees) between the sprockets. The upwards vertical force to the right of the center of the wheel axis generates a counter clockwise torque, but the tension force in the chain will be countered by the compression forces in the framework, so the linear forces cancel.

To compensate for the offset tension (upwards force on the wheel axis), I'm thinking that the compression force in the right post will be greater than the left post, which makes sense in that this would generate a lateral force to the left on the hub, corresponding to the counter-clockwise torque on the vehicle.

I'm also thinking that the wheel axis could have been extended equally in both directions, and would stil provide the same roll prevention as just extending the left axis.

 

[OmCheeto]

I got that, but take the case where the right side of the prop sprocket is directly above the front of wheel axis sprocket, so that the chain under tension takes a vertical run (just twisting 90 degrees) between the sprockets. The upwards vertical force to the right of the center of the wheel axis generates a counter clockwise torque, but the tension force in the chain will be countered by the compression forces in the framework, so the linear forces cancel.

To compensate for the offset tension (upwards force on the wheel axis), I'm thinking that the compression force in the right post will be greater than the left post, which makes sense in that this would generate a lateral force to the left on the hub, corresponding to the counter-clockwise torque on the vehicle.

I'm also thinking that the wheel axis could have been extended equally in both directions, and would stil provide the same roll prevention as just extending the left axis.

I will probably replace the chain with a set of gears in my analysis. Chains are too much like noodles, and I have no idea how to create a free body diagram with a twisted noodle.

 

[ThinAirDesign]

I'm also thinking that the wheel axis could have been extended equally in both directions, and would stil provide the same roll prevention as just extending the left axis.

Of course if we had extended the right side the same length we extended the left side we would have had the same roll protection (actual marginally more because of the added weight of the longer right side). Problem is we would have no longer been able to trailer the Blackbird legally down the road as it would been too wide.

Extending only the left side gave us 99% of the increase (as compared to both sides) while not adding the disadvantages. Extending the right side would have essentially been a waste of time.

JB

 

[rcgldr]

Of course if we had extended the right side the same length we extended the left side we would have had the same roll protection (actual marginally more because of the added weight of the longer right side). Problem is we would have no longer been able to trailer the Blackbird legally down the road as it would been too wide.

That's what I thought about the roll protection, but I wasn't aware of the trailering issue. Thanks for the response.

 

[Llyricist]

Okay, someone wanted math?

I'll take a cart that is very much like the actual cart, weighs 650 lbs (295 kg) w/pilot;

Has fairly efficient wheels giving a coefficient of rolling resistance (crr) of .013;

The fairings and such give it a relatively sleek coefficient of drag of .26 and it has a frontal area of 20 Sq Ft, or 1.858 m²;

It uses a twisted chain drive transmission with a 23 tooth sprocket on the axle, and a 65 tooth sprocket on the propeller shaft, and the twist will take the normal 98% efficiency of a chain drive down to 92%;

It has a propeller designed and modeled in Javaprop based on the geometry and airfoils specified by spork elsewhere, with a diameter of 17.5 feet (5.33 m);

And one more specification is that the wheels have an outer diameter of 27 inches (.686 m).


Now I'm just going to analyse the main forces acting on the cart when it is travilling at 24 MPH (10.72 m/s) with a 12 MPH (5.36 m/s) tailwind. That's twice wind speed. Additionally I will use the conditions at El Mirage Lake Bed on July 3, 2010 to derive an air density of 1.063 kg/m³. All the caluculations will be done using SI units for convenience.

First, there will be a rolling resistance acting on the cart of (crr * mass * 1G) 0.013 * 295 kg * 9.81 = 37.6 N

Next there will be aerodynamic drag, using the Rayleigh formula: (F_D = 0.5 * density * v² * coeeficient of Drag * frontal area);

0.5 * 1.063 kg/m³ * 5.36 m/s (airspeed: 10.72 m/s - 5.36 m/s) * .26 * 1.858 m² = 7.39 N

Next, using the gearing and wheel size specified above, I determine that while the wheels are rolling along the ground at 10.72 m/s the prop shaft will be turning at 105.7 rpm.

Plugging that, an airspeed of 5.36 m/s and density of 1.063 kg/m³ into javaprop tells me the propeller will demand a torque of 270.38 N-m.

Transmitting that torque through the gears to the axle will result in an opposing torque on the axle of 270.38 * (23 / 65) / .92 (drive efficiency) = 104 N-m

Dividing that by the radius of the wheel will result in an opposing force of 104 N-m / .343 m = 303.2 N.

For total opposing Forces of: 37.6 N + 7.39 N + 303.2 N = 348.2 N

Javaprop also gives a thrust in those conditions of 376.6 N, For a net acceleration force of 376.6 N - 348.2 N = 28.4 N;

Which will accelerate the cart at a rate of (a = F / m) 28.4 N / 295 kg = 0.09 m/s/s, well in line with the observed and documented performance of the actual cart.

Next I will quantify how the energy balances in a frame of reference where the ground can be considered stationary.

 

[OmCheeto]

Okay, someone wanted math?

I'll take a cart that is very much like the actual cart, weighs 650 lbs (295 kg) w/pilot;

Has fairly efficient wheels giving a coefficient of rolling resistance (crr) of .013;

The fairings and such give it a relatively sleek coefficient of drag of .26 and it has a frontal area of 20 Sq Ft, or 1.858 m²;

It uses a twisted chain drive transmission with a 23 tooth sprocket on the axle, and a 65 tooth sprocket on the propeller shaft, and the twist will take the normal 98% efficiency of a chain drive down to 92%;

It has a propeller designed and modeled in Javaprop based on the geometry and airfoils specified by spork elsewhere, with a diameter of 17.5 feet (5.33 m);

And one more specification is that the wheels have an outer diameter of 27 inches (.686 m).


Now I'm just going to analyse the main forces acting on the cart when it is travilling at 24 MPH (10.72 m/s) with a 12 MPH (5.36 m/s) tailwind. That's twice wind speed. Additionally I will use the conditions at El Mirage Lake Bed on July 3, 2010 to derive an air density of 1.063 kg/m³. All the caluculations will be done using SI units for convenience.

First, there will be a rolling resistance acting on the cart of (crr * mass * 1G) 0.013 * 295 kg * 9.81 = 37.6 N

Next there will be aerodynamic drag, using the Rayleigh formula: (F_D = 0.5 * density * v² * coeeficient of Drag * frontal area);

0.5 * 1.063 kg/m³ * 5.36 m/s (airspeed: 10.72 m/s - 5.36 m/s) * .26 * 1.858 m² = 7.39 N

Next, using the gearing and wheel size specified above, I determine that while the wheels are rolling along the ground at 10.72 m/s the prop shaft will be turning at 105.7 rpm.

Plugging that, an airspeed of 5.36 m/s and density of 1.063 kg/m³ into javaprop tells me the propeller will demand a torque of 270.38 N-m.

Transmitting that torque through the gears to the axle will result in an opposing torque on the axle of 270.38 * (23 / 65) / .92 (drive efficiency) = 104 N-m

Dividing that by the radius of the wheel will result in an opposing force of 104 N-m / .343 m = 303.2 N.

For total opposing Forces of: 37.6 N + 7.39 N + 303.2 N = 348.2 N

Javaprop also gives a thrust in those conditions of 376.6 N, For a net acceleration force of 376.6 N - 348.2 N = 28.4 N;

Which will accelerate the cart at a rate of (a = F / m) 28.4 N / 295 kg = 0.09 m/s/s, well in line with the observed and documented performance of the actual cart.

Next I will quantify how the energy balances in a frame of reference where the ground can be considered stationary.

Cool. I thought I was going to have to do all the maths.

Given that this thread is a 6th generation ddw...'er, and I've seen what seems to have been about 10,000 posts on this topic, and seeings how you joined the forum today, and seem really familiar with Spork's posts:

It has a propeller designed and modeled in Javaprop based on the geometry and airfoils specified by spork elsewhere, with a diameter of 17.5 feet (5.33 m);

...


Plugging that, an airspeed of 5.36 m/s and density of 1.063 kg/m³ into javaprop tells me the propeller will demand a torque of 270.38 N-m.

Can you give me either the link, here at the forum, or the numbers themselves, for the propeller.

I tried getting "javaprop" a few minutes ago, but whatever site I ended up at, seemed to be trying to give me some kind of computer virus, and I had to terminate the connection.



I know the truth is out there...

https://www.youtube.com/watch?v=JDZBgHBHQT8

 

[Llyricist]

Cool. I thought I was going to have to do all the maths.

Given that this thread is a 6th generation ddw...'er, and I've seen what seems to have been about 10,000 posts on this topic, and seeings how you joined the forum today, and seem really familiar with Spork's posts:


Can you give me either the link, here at the forum, or the numbers themselves, for the propeller.

Here's the numbers I used:

r(ft) chord(ft) incidence(deg)
0.93983 0.18983 80.1249
1.06718 0.31718 76.89058
1.21091 0.46091 73.75922
1.36252 0.61252 70.74371
1.51452 0.76452 67.8535
1.66077 0.91077 65.09468
1.7967 1.0467 62.47044
1.91922 1.16922 59.98132
2.02652 1.27652 57.62588
2.11789 1.36789 55.40093
2.1934 1.4434 53.30211
2.25373 1.50373 51.32415
2.29988 1.54988 49.46123
2.33312 1.58312 47.70721
2.35476 1.60476 46.05583
2.36616 1.61616 44.50084
2.3686 1.6186 43.03614
2.3633 1.6133 41.65586
2.35135 1.60135 40.35435
2.33373 1.58373 39.12632
2.31133 1.56133 37.96673
2.28491 1.53491 36.87091
2.25513 1.50513 35.83446
2.22255 1.47255 34.85332
2.18766 1.43766 33.92371
2.15088 1.40088 33.04211
2.11254 1.36254 32.20531
2.07293 1.32293 31.41029
2.03228 1.28228 30.65429
1.99078 1.24078 29.93474
1.94858 1.19858 29.24929
1.90579 1.15579 28.59575
1.86249 1.11249 27.97211
1.81875 1.06875 27.37649
1.77457 1.02457 26.80716
1.72997 0.97997 26.26254
1.68491 0.93491 25.74113
1.63934 0.88934 25.24157
1.59319 0.84319 24.76258
1.54632 0.79632 24.30298
1.49857 0.74857 23.86167
1.44974 0.69974 23.43764
1.39951 0.64951 23.02992
1.34748 0.59748 22.63764
1.29305 0.54305 22.25996
1.23533 0.48533 21.89613
1.1729 0.4229 21.54542
1.10312 0.35312 21.20715
1.02023 0.27023 20.8807
0.90414 0.15414 20.56549
1.10312 0.35312 21.20715
1.02023 0.27023 20.8807
0.90414 0.15414 20.56549

You can copy just the numbers to your clipboard and when you bring up the import geometry dialog in Javaprop, it will paste them automatically.

Also, the airfoils used are not included in javaprop, I had to follow the directions and used the same guy's JavaFoil program to create the necessary data files for the NACA 6412.


Also also, I learned java and accessed the "propeller object" directly from my own program to get more specific results than are provided by the program the way it is.

I tried getting "javaprop" a few minutes ago, but whatever site I ended up at, seemed to be trying to give me some kind of computer virus, and I had to terminate the connection.

was it this site?

http://www.mh-aerotools.de/airfoils/javaprop.htm"

If so, that is surprising, I never had any propblem

 

[Llyricist]

The first thing about the energy balance to understand is that the external input of energy to the cart system only has to equal the losses in the system plus any gain in KE of the system.

The losses of the cart system (again, all in a frame of reference where the ground can be considered stationary) are the aerodynamic drag, the rolling resistance, the drivetrain losses, and the aerodynamic losses at the propeller. It is important to note here that the negative work of the prop on the cart is not a loss, since much of it is put right back into the system, just the part lost in the drivetrain and aerodynamically are losses.

And the cart will gain KE as it accelerates from 10.72 to 10.81 m/s over the course of a second.


So starting with the aerodynamic drag, that is a force of 7.39 N acting over 10.72 m/s, for a loss to the system of 10.72 * 7.39 = 79.2 J/s.

Next the rolling resistance, a force of 37.6 N acting over 10.72 m/s gives a loss of 403.1 J/s.

Now it starts to get more complicated. The torque demand at the propeller shaft equates to a power demand of 2993.5 J/s (Watts), While the negative work on the cart from that demand equals 303.2 N * 10.72 m/s = 3250.3 J/s, giving a drivetrain loss of 3250.3 - 2993.5 = 256.8 J/s.

So far, that is 79.2 + 403.1 + 256.8 = 739.1 J/s in losses.

Skipping the aerdynamic losses at the prop for the moment, since the same calculations will be used to determine the energy input, The ke gain of the cart system can be calculated from the difference in KE after and before.

KE = 0.5 * m * v²

before = 0.5 * 295 kg * 10.72 m/s² = 16950.46 J
after = 0.5 * 295 kg * 10.81 m/s² = 17236.27 J

Difference = 285.8 J

So far, an energy requirement of 739.1 + 285.8 = 1024.9 J

Now for the really complicated part.

From http://www.mh-aerotools.de/airfoils/propuls4.htm" , which is the same site that Javaprop comes from, there is a generic formula for propeller thrust:

Where;

T = thrust [N]
D = propeller diameter [m]
v = velocity of incoming flow [m/s]
Δv = additional velocity, acceleration by propeller [m/s]
ρ = density of fluid [kg/m³]

Since I already have everything except Δv, I have to solve for that, which results in a quadratic equation that is too messy to post unless absolutely necessary. But I need that value both for the KE loss of the air and the prop's aerodynamic losses.

Long story short, I get a Δv of 2.417 m/s.

With that in hand I move to the "Engine Power" formula on that same page, to get the minimum power, assuming no losses at all, for a propeller to generate that thrust in those conditions.

Which results in a power of 2473.6 W, which is far less than the 2993.5 W that is actually at the shaft. In fact, the difference is the aerodynamic losses at the prop.

So those losses are 2993.5 - 2473.6 = 519.9 J/s

For total losses and KE gain of the cart system of 519.9 + 1024.9 = 1544.8 J/s


Now, for the KE loss of the air that is actually transferred to the cart, I need the mass of the air that is decelerated, and how fast it was going before and after.

The thrust formula above is actually F = ma, where everything on the right side except for the last Δv is the mass (it calculates the volume of the cylinder of air and multiplies by density), and the last Δv is the acceleration.

So using that part to calculate the mass results in a mass of 155.79 kg.

The KE of the air before the cart moves through it is 0.5 * 155.79 kg * 5.36² = 2238 J

The air is decelerated by 2.417 m/s, so its velocity after the cart moves through it is 5.36 - 2.417 = 2.94 m/s;

so its KE will now be 0.5 * 155.79 kg * 2.94² = 673.3 J

For an energy input to the cart system of 2238 - 673.3 = 1564.7 J/s


So we have 1564.7 J/s into the system, and 1544.8 either lost, or gained in KE, by the cart system. Close enough considering all the rounding I did, and the lack of using calculus for really precise answers.

 

[RonL]

Cool. I thought I was going to have to do all the maths.

Given that this thread is a 6th generation ddw...'er, and I've seen what seems to have been about 10,000 posts on this topic, and seeings how you joined the forum today, and seem really familiar with Spork's posts:


Can you give me either the link, here at the forum, or the numbers themselves, for the propeller.

I tried getting "javaprop" a few minutes ago, but whatever site I ended up at, seemed to be trying to give me some kind of computer virus, and I had to terminate the connection.



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