# Forget planes on treadmills, here's a sail-cart out-sailing the wind! **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [November 7, 2010, 8:17am UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778 "2010-11-07T08:17:09Z")\ **Posts on this page:** 20\ **Page:** 3
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [November 9, 2010, 6:56pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/41 "2010-11-09T18:56:24Z")
> [@hdc\_bst](#): > > What formulation of work are you using that is dependent on the velocity? We can consider work as a change of energy or evaluate work done by a mechanical force (force x distance). In either case, terms may exist which are completely independent of instantaneous velocity. In the problem at hand, the best work analysis is a torque applied through a revolution No, we can not consider work as a change of energy because thats not what it is. Work is the **transfer** of energy. If there is no energy transfer, then there is no work being done. Using your definition, if I throw a ball against the wall, the wall has done work because the balls energy has changed. That is just wrong. I am being a bit sloppy with my work/power by using velocity. Since V=0, X is also zero. So in any event, Power=Force_Velocity or Work=Force_Distance and both are zero. > [@hdc\_bst](#): > > In the end the energy does come from the wind, but the ground is used to convert it. I understand this, but saying things like the power to turn the propellor comes from the ground is just wrong. > [@hdc\_bst](#): > > It is moving energy (work) from a larger system, the entire craft’s kinetic energy, to a smaller system, the massless axle and generator. From the generator’s perspective, the ground is doing work (by applying a torque). From the generator’s reference frame yes, but we aren’t in the generator’s reference frame. We are in the reference frame of the earth or ground. In that reference frame, there is no work being done by the ground, because there is no energy transfer from the ground to the system. > [@ThinAirDesigns](#): > > treis – let’s get practical for a moment and leave the theoretical behind. > > If we place a rubber tire linked directly to a generator on the end of an arm and lower the tire to the pavement from a steadily propelled platform, can generate electricity? > > Thanks > > JB Yes, but there is no work done by the ground because there is no energy transfer from the ground to the system. In the reference frame of the earth, as I said previously, the velocity or delta X is zero at the application of the force. Therefore there is no work or power being done by the ground. The power to turn the generator is coming from the force you are applying along with the velocity of the car. ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 9, 2010, 7:52pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/42 "2010-11-09T19:52:17Z")
Me: > [@](#): > > If we place a rubber tire linked directly to a generator on the end of an arm and lower the tire to the pavement from a steadily propelled platform, can generate electricity? > [@treis](#): > > Yes, but there is no work done by the ground because there is no energy transfer from the ground to the system. If you would humor me for just a bit I really would like you and I to leave the ‘what does work on what’ theory behind for a bit and just focus on what _works_. Thanks. So you agree with my premise that given a steadily propelled platform, using a tire rolling across the ground hooked to a generator to produce electricity _actually works_. Excellent. Let’s see if we can come to an agreement on just how much electricity we can produce based on the force exerted on our ‘steadily propelled platform’. First, let’s make sure we agree on the definitions: A: 1.0 HP = 550 foot-pounds per second 1.0 HP = 750 watts ([Horsepower - Wikipedia](http://en.wikipedia.org/wiki/Horsepower)) ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 9, 2010, 7:58pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/43 "2010-11-09T19:58:13Z")
Me: > [@](#): > > If we place a rubber tire linked directly to a generator on the end of an arm and lower the tire to the pavement from a steadily propelled platform, can generate electricity? > [@treis](#): > > Yes, but there is no work done by the ground because there is no energy transfer from the ground to the system. If you would humor me for just a bit I really would like you and I to leave the ‘what does work on what’ theory behind for a bit and just focus on what _works_. Thanks. So you agree with my premise that given a steadily propelled platform, using a tire rolling across the ground hooked to a generator to produce electricity _actually works_. Excellent. Let’s see if we can come to an agreement on just how much electricity we can produce based on the force exerted on our ‘steadily propelled platform’. (all losses ignored for the moment) Using the below definitions(with reference link), can we agree that if our platform is moving at 55ft per second and is pulled along by an available force of 10lbs, that we can load up the generator such it will produce 750 watts of electrical power for us? Thanks. 1.0 HP = 550 foot-pounds per second 1.0 HP = 750 watts ([Horsepower - Wikipedia](http://en.wikipedia.org/wiki/Horsepower)) ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 9, 2010, 8:01pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/44 "2010-11-09T20:01:02Z")
Treis, I apologize that somehow I submitted a post before I was through with it and now can’t delete it. Please respond to post #43 if you will and ignore the (admittedly nearly identical) first one. Thanks. JB ---
**Author:** ![hdc\_bst](https://avatars.discourse-cdn.com/v4/letter/h/9f8e36/32.png) [@hdc\_bst](https://boards.straightdope.com/u/hdc_bst)\ **Post date:** [November 10, 2010, 12:22am UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/45 "2010-11-10T00:22:26Z")
Once more into the brink. **treis** I would be interested to know if you believe: a) DDWFTTW is a sham, and the NALSA was duped / is complicit b) DDWFTTW is real, but the explanations in this thread are all wrong EDIT: I’m now a believer in b) if we restrict b) to my earlier explanations. Now to the direct quotes: > [@treis](#): > > No, we can not consider work as a change of energy because thats not what it is. Work is the **transfer** of energy. If there is no energy transfer, then there is no work being done. Using your definition, if I throw a ball against the wall, the wall has done work because the balls energy has changed. That is just wrong. Actually that is correct. A transfer of energy results in a change of energy. Delta E. If you throw a ball against a wall there is either: no change in energy due to a perfectly elastic collision (the wall does no work), or a change in energy due to an inelastic collision, in which case work is done: the wall applies a force to the ball over a distance due to the ball’s deformation. > [@](#): > > I understand this, but saying things like the power to turn the propellor comes from the ground is just wrong. No. It’s perfectly appropriate to mention any intermediary we would like in our energy transition. Like the Man of Steel, I derive my energy from our yellow sun. However, it is just as appropriate to say that the energy to power me comes from plants. EDIT: This was just such a beautiful piece of snark, I’m leaving it in, but boy is my face red. On review, the error is revealed if I change the analogy to say that the energy to power me comes from poo. > [@](#): > > From the generator’s reference frame yes, but we aren’t in the generator’s reference frame. Why not? There is no privileged frame of reference. But we can choose another if you’d prefer. > [@](#): > > We are in the reference frame of the earth or ground. In that reference frame, there is no work being done by the ground, because there is no energy transfer from the ground to the system. And here I need to issue an apology - we may be close to converging on understanding. I was working on a nice, semi-snarky reply, but during it I discovered that I nearly agree with you - there is no work being done by the ground - there is work being done to the ground. What is happening from the ground’s perspective? At t = 0, something is in contact with the ground at point x = 0. (This something is a point on a wheel). Due to static friction this something is applying a force on the ground in direction = 1; thanks to Newton the ground knows to push back with an equal force in direction = -1. We know that the force involved is not zero because the wheel turns. A short time later, t = 0 + e, and something is applying a force on the ground at point x = 0 + v_e. The forces are similar. For a whole bunch of these little instances, something is applying a force on the ground along a linear path from x = 0 to x = x1. And so, depending on what that force is, work is done to the ground by the something: W = F_x1. We can also consider that work is being done to our axle system, to the tune of W = (r_F) \* omega. For one revolution of the axle, W-to-the-ground = F_2_pi_r, W-to-the-axle = F_2_pi\*r. Where does this energy come from? As you previously maintained, it comes from the doofus pushing the device / the wind. Keeping our notional generator attached, the wind exerts a force equal to twice the back electromotive force. Limiting factors include maximum force the wind can apply (aerodynamics question) and the maximum static friction your wheel/ground interface can maintain. We can assume that energy available to the wind is virtually limitless. If you then use your available energy to power a fan, you can push back against the wind, making it push on you even harder, sending you speeding down wind faster than the wind. There are probably limits here based on what your spinning propellor/fan does to your effective cross section into the wind. Aerodynamics makes my head hurt. > [@](#): > > Yes, but there is no work done by the ground because there is no energy transfer from the ground to the system. In the reference frame of the earth, as I said previously, the velocity or delta X is zero at the application of the force. Therefore there is no work or power being done by the ground. The power to turn the generator is coming from the force you are applying along with the velocity of the car. Yep. Gotta go with your analysis here. The energy transfer is from the wind - to the system - partially to the ground. The neat thing is that the work done to the ground gives you something to ‘push off against’ letting you extract more work from that lazy, lazy wind. Again, my apologies for the earlier snark. It works, but not the way I initially thought it did. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [November 10, 2010, 3:27pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/46 "2010-11-10T15:27:05Z")
> [@ThinAirDesigns](#): > > Me: > > If you would humor me for just a bit I really would like you and I to leave the ‘what does work on what’ theory behind for a bit and just focus on what _works_. Thanks. > > So you agree with my premise that given a steadily propelled platform, using a tire rolling across the ground hooked to a generator to produce electricity _actually works_. Excellent. > > Let’s see if we can come to an agreement on just how much electricity we can produce based on the force exerted on our ‘steadily propelled platform’. > > (all losses ignored for the moment) > > Using the below definitions(with reference link), can we agree that if our platform is moving at 55ft per second and is pulled along by an available force of 10lbs, that we can load up the generator such it will produce 750 watts of electrical power for us? > > Thanks. > > 1.0 HP = 550 foot-pounds per second > 1.0 HP = 750 watts > ([Horsepower - Wikipedia](http://en.wikipedia.org/wiki/Horsepower)) This is correct. The point is that if we use the ground does work concept, you would need to add in that work to the analysis, which would result in an incorrect answer. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [November 10, 2010, 3:41pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/47 "2010-11-10T15:41:41Z")
> [@hdc\_bst](#): > > Once more into the brink. **treis** I would be interested to know if you believe: > > a) DDWFTTW is a sham, and the NALSA was duped / is complicit > b) DDWFTTW is real, but the explanations in this thread are all wrong > > EDIT: I’m now a believer in b) if we restrict b) to my earlier explanations. B. > [@hdc\_bst](#): > > Actually that is correct. A transfer of energy results in a change of energy. Delta E. If you throw a ball against a wall there is either: no change in energy due to a perfectly elastic collision (the wall does no work), or a change in energy due to an inelastic collision, in which case work is done: **the wall applies a force to the ball over a distance** due to the ball’s deformation. Really? The wall moves? > [@hdc\_bst](#): > > Why not? There is no privileged frame of reference. But we can choose another if you’d prefer. There is no privileged frame of reference, in the sense that you can do the math using any frame. There is a privileged frame of reference in the sense that we only care about what happens in the reference frame of the earth. > [@hdc\_bst](#): > > And here I need to issue an apology - we may be close to converging on understanding. I was working on a nice, semi-snarky reply, but during it I discovered that I nearly agree with you - there is no work being done by the ground - there is work being done to the ground. No there isn’t, at least in the reference frame of the ground. > [@hdc\_bst](#): > > What is happening from the ground’s perspective? At t = 0, something is in contact with the ground at point x = 0. (This something is a point on a wheel). Due to static friction this something is applying a force on the ground in direction = 1; thanks to Newton the ground knows to push back with an equal force in direction = -1. We know that the force involved is not zero because the wheel turns. A short time later, t = 0 + e, and something is applying a force on the ground at point x = 0 + v_e. The forces are similar. For a whole bunch of these little instances, something is applying a force on the ground along a linear path from x = 0 to x = x1. And so, depending on what that force is, work is done to the ground by the something: W = F_x1. No, the force is always applied where V=0 and thus there is never any change in distance at the point of contact. No delta x means no work. > [@hdc\_bst](#): > > We can also consider that work is being done to our axle system, to the tune of W = (r_F) \* omega. For one revolution of the axle, W-to-the-ground = F_2_pi_r, W-to-the-axle = F_2_pi\*r. No you can’t. Doing so will result in an incorrect solution because you must also account for the force on the axle over a linear distance. ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 10, 2010, 4:44pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/48 "2010-11-10T16:44:01Z")
Me: > [@](#): > > … can we agree that if our platform is moving at 55ft per second and is pulled along by an available force of 10lbs, that we can load up the generator such it will produce 750 watts of electrical power for us? > [@treis](#): > > This is correct. Thanks treis. Would you also agree with the following: 1/2hp = 275 foot-pounds per second 1/2hp = 375 watts This means that at 27.5ft/sec, 10lbs of force can be produced by a lossless propeller consuming 1/2hp (375 watts). Thanks. JB ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [November 10, 2010, 6:53pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/49 "2010-11-10T18:53:38Z")
Look, I’m not going to play answer 20 questions. I understand perfectly the underlying principal of using the wheels to turn the propeller. It is essentially no different from the compressor on a jet engine. The problem is that the wording is sloppy. I would be more interested in hearing about the principles behind the propeller producing more energy with a tailwind. ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 10, 2010, 7:04pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/50 "2010-11-10T19:04:35Z")
> [@treis](#): > > Look, I’m not going to play answer 20 questions. Tries - there is no “20 question” attempt here, there is only an attempt to understand what you mean by the following: > [@treis](#): > > …anyone claiming that the wheels are driving the prop resulting in acceleration is either (1) crappy at explaining things or (2) full of shit. It is a simple fact that in the Blackbird, at all speeds the torque to drive the propeller is coming from the wheels. It is a fact that thrust of the propeller results in the acceleration of the vehicle. Those two facts appear to conflict with your statement. Clearly the path I have taken in trying to understand it is frustrating you – perhaps you could explain in different words and we could go from there. Thanks JB ---
**Author:** ![hdc\_bst](https://avatars.discourse-cdn.com/v4/letter/h/9f8e36/32.png) [@hdc\_bst](https://boards.straightdope.com/u/hdc_bst)\ **Post date:** [November 11, 2010, 12:04am UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/51 "2010-11-11T00:04:35Z")
> [@treis](#): > > B. Ok, good. And thanks for sticking around, this has been a fun exercise, and your obstinate practice of being right has helped me work through a better understanding. > [@](#): > > Really? The wall moves? Not if it has infinite mass. It is a necessary catalyst for the transformation of energy as a portion of the ball’s kinetic energy is used to deform and heat the ball. > [@](#): > > There is no privileged frame of reference, in the sense that you can do the math using any frame. There is a privileged frame of reference in the sense that we only care about what happens in the reference frame of the earth. I think that the car or wind are also interesting frames of reference. > [@](#): > > No there isn’t, at least in the reference frame of the ground. But, but, but… Thanks for pointing this out. I think this is where the conveyor belt hurt my understanding. If in the conveyor belt example we set our frame of reference on the belt itself, in that frame the belt does no work on the wheel. So, in the end, the ground is a necessary component of the energy transfer but does no work (from the perspective of an observer on the ground). It does induce a rotation in the wheel. This means that the energy from the wind can be converted into rotational energy/potential energy harvested from the generator. The trick seems to be that the wind is capable of providing a large amount of energy. As the frictional (rolling) force increases, the force delivered by the wind to propel an object at wind speed also increases, and work done by the wind increases. The force on a sail is proportional to apparent wind speed. With our generator running we should travel at slightly less than wind speed. What I’m stuck on now is how this is sustainable when the propeller is turned on, which seems like it would result in an apparent wind speed against the direction of travel. Maybe the way to visualize it is that air moved by the propeller does work on the vehicle in the direction of travel, and that the body of wind in general does work on the air displaced by the propeller. I’m a bit more confused now, although I’ve finally managed to wrap my mind around rotational friction not doing work. ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 11, 2010, 1:50pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/52 "2010-11-11T13:50:45Z")
Treis, it’s becoming pretty clear that you’ve just backed yourself in a corner here and just want your silly statement to go away through your silence. Since I’m the guy you said was either ‘crappy’ or ‘full of shit’, I’m not going to let you just slink off the topic quite so easy. Here was my statement regarding the wheels turning the propeller: **A:** > [@ThinAirDesigns](#): > > – the wheels provide the torque to turn the rotor (always). > – the rotor does not provide the torque to turn the wheels (ever). Here was what you said about it: **B:** > [@treis](#): > > … anyone claiming that the wheels are driving the prop resulting in acceleration is either (1) crappy at explaining things or (2) full of shit. And later when I try to walk you through to show that you’re wrong: C: > [@treis](#): > > Look, I’m not going to play answer 20 questions. I understand perfectly the underlying principal of using the wheels to turn the propeller. … The problem is that the wording is sloppy. Just man up and admit that you’ve realized that your statement “B” is wrong. My original statement is true as you’ve now admitted in “C”, is not flawed in explanation in any way. Heres my favorite though: Please support your assertion that the “wording is sloppy” in the following: > [@](#): > > “-- the wheels provide the torque to turn the rotor (always).” Yeah, I’m sure you’ll get right on that task. :rolleyes: JB ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [November 11, 2010, 2:19pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/53 "2010-11-11T14:19:05Z")
Please give me your infinite forgiveness for taking a whole 20 hours to respond to a post. > [@ThinAirDesigns](#): > > It is a simple fact that in the Blackbird, at all speeds the torque to drive the propeller is coming from the wheels. It is a fact that thrust of the propeller results in the acceleration of the vehicle. Those two facts appear to conflict with your statement. If the thrust of the propeller resulted in the acceleration than it would work regardless of whether it had a tailwind. Leaving out the tailwind, the most important part, makes it a bad explanation. ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 11, 2010, 4:00pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/54 "2010-11-11T16:00:28Z")
> [@treis](#): > > Leaving out the tailwind, the most important part, makes it a bad explanation. Yeah, that’s all well and good exept for the fact that I \*\*DIDN’T \*\*“leave at the tailwind”. Again from my FIRST post with the tailwind portions emphasized: > [@](#): > > Here is the claim: “**Directly downwind[**, faster than the wind, \*\*powered only by the wind’\*\*steady state”. > > basics: > > – by design, this particular one works best when aimed **directly downwind**. > > – a simple gearing change will cause it to go directly upwind rather than **down** And now for your next lousy excuse for being snotty and wrong at the same time ?? JB ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 11, 2010, 4:37pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/55 "2010-11-11T16:37:45Z")
> [@treis](#): > > There is a privileged frame of reference in the sense that we only care about what happens in the reference frame of the earth. > [@hdc\_bst](#): > > I think that the car or wind are also interesting frames of reference. Yes hdc, it’s just nonsense to make the assertion that one frame is privileged because “we only care about the earth frame”. This is the domain of folks who come in throwing BS around like: > [@treis](#): > > … if velocity at the point of contact is 0, is that there is no possibility of net work being done by the friction force. This is high school physics. … I whole heartedly deny that, using the normal assumptions of friction, the ground is doing work on the car. As everyone here knows, work and energy are frame dependent. Treis saying that the ground isn’t doing work on the car simply becasue _HE_ claims to only care about the ground frame doesn’t make him any more correct. The wheels _always_ provide the torque to turn the propeller – _frame independent_. Depending on the frame of reference chosen (and as you point out there are excellent and instructive reasons to look at things from other frames), the power and work can come either from the wind, or from the ground. That’s _correct_ high school physics. JB ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 11, 2010, 4:47pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/56 "2010-11-11T16:47:59Z")
> [@hdc\_bst](#): > > The force on a sail is proportional to apparent wind speed. I’m pretty sure you know this and have used more casual reference, but actually it would be more accurate to say that the force on the sail goes up proportional to the _square_ of the apparent wind speed. > [@](#): > > What I’m stuck on now is how this is sustainable when the propeller is turned on, … It’s sustainable because the vehicle is moving across the ground faster than it is moving through the air and thus the wheels can produce more energy than the prop needs to sustain. (see my full post @#20) > [@](#): > > … which seems like it would result in an apparent wind speed against the direction of travel. If I understand what you are saying here, once the vehicle accelerates past wind speed, the apparent wind _is_ against the direction of travel. This apparent headwind must be overcome to sustain or accelerate and ultimately is one of the losses that limits the top speed of the craft. > [@](#): > > Maybe the way to visualize it is that air moved by the propeller does work on the vehicle in the direction of travel, and that the body of wind in general does work on the air displaced by the propeller. Perhaps this is just semantics, but it might be more accurate to say that “the body of wind in general _has already done_ work on the air displaced by the propeller”. After all, air the propeller is interacting with had already been accelerated to wind speed long before we got to it. JB ---
**Author:** ![Redpill](https://avatars.discourse-cdn.com/v4/letter/r/58956e/32.png) [@Redpill](https://boards.straightdope.com/u/Redpill)\ **Post date:** [November 11, 2010, 5:54pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/57 "2010-11-11T17:54:12Z")
> [@ThinAirDesigns](#): > > It is a simple fact that in the Blackbird, at all speeds the torque to drive the propeller is coming from the wheels. It is a fact that thrust of the propeller results in the acceleration of the vehicle. Those two facts appear to conflict with your statement. When the vehicle is at same speed as wind, to go even faster, something has to start turning faster first to affect other. If wheels start turning faster, how? If propeller starts turning faster, how? Maybe the gearing can use the momentum to turn propeller faster, but that will cause braking effect. ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 11, 2010, 7:18pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/58 "2010-11-11T19:18:08Z")
> [@Redpill](#): > > When the vehicle is at same speed as wind, to go even faster, something has to start turning faster first to affect other. If wheels start turning faster, how? If propeller starts turning faster, how? Since both the prop and wheels are geared together using a fixed ratio transmission, one cannot start turning faster without the other also turning proportionally faster as well. This precludes your opening statement above from being technically true. I suspect however that this is just a minor wording issue on your part as the 2 questions following your statement are reasonable ones. Since they are tied together, it initially becomes an issue of force, rather than one of speed. At below wind speed, if the force of any tailwind drag combined with any forward force (thrust) of the propeller is greater than the retarding force at the wheels, the vehicle will accelerate and both the wheels and propeller will turn faster. Exactly at wind speed and above wind speed, if the forward force (thrust) of the propeller is greater than any headwind drag combined with the retarding force at the wheels, the vehicle will accelerate and both the wheels and propeller will turn faster. So, the answer to your ‘chicken or egg’ question that opens this post is that at exactly wind speed, the forward force of the propeller on the Blackbird is greater than the retardant force of the wheels and thus the vehicle accelerates and both the prop and wheels turn faster. My previous post (#20) goes into the energy and force balance issues underlying the above answer. Happy to answer more questions as well or delve deeper if you wish. > [@](#): > > Maybe the gearing can use the momentum to turn propeller faster, but that will cause braking effect. As the Blackbird accelerates, every component of the vehicle gains momentum. At no time is there any momentum in one part of the vehicle being exchanged to make another part spin faster – it simply isn’t possible in the Blackbird with a fixed ratio transmission. ---
**Author:** ![Redpill](https://avatars.discourse-cdn.com/v4/letter/r/58956e/32.png) [@Redpill](https://boards.straightdope.com/u/Redpill)\ **Post date:** [November 11, 2010, 7:35pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/59 "2010-11-11T19:35:26Z")
Thanks for taking the time to answer, I’m bad at physics. At the moment wind speed is reached, what causes the propeller to spin even faster? ---
**Author:** ![ThinAirDesigns](https://avatars.discourse-cdn.com/v4/letter/t/df705f/32.png) [@ThinAirDesigns](https://boards.straightdope.com/u/ThinAirDesigns)\ **Post date:** [November 11, 2010, 7:53pm UTC](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778/60 "2010-11-11T19:53:20Z")
> [@Redpill](#): > > Thanks for taking the time to answer, I’m bad at physics. No problem Red. > [@](#): > > At the moment wind speed is reached, what causes the propeller to spin even faster? It’s the imbalance of forces that I described above: “… at exactly wind speed, the forward force of the propeller on the Blackbird is greater than the retardant force of the wheels and thus the vehicle accelerates and both the prop and wheels turn faster.” The root cause of the imbalance of course is the true tailwind which continues to blow unabated across the ground, even while the vehicle chassis itself is traveling at exactly wind speed and thus sees an apparent wind of zero. JB [Previous page](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778.md?page=2) [Next page](https://boards.straightdope.com/t/forget-planes-on-treadmills-heres-a-sail-cart-out-sailing-the-wind/559778.md?page=4)