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Merged DDWFTTW in a “Real” Wind

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humber said:
The vane stalled, and that is can be easily calculated...but not by you.
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You have no idea what you are talking about. I spotted the problem with the data from just the record run, which is why I took the time to figure out the offset, and recalculate the wind direction and speed while the CV was stationary by using a "best guess" heading from when the CV started moving in order to tame the results for the graphs I made.

My graphs sitll show a shift when the CV starts accelerating, but that's because I used a heading from too early in its acceleration, which was still off, for reasons that have been explained by many already.

I also spotted the problem when the CV accelerated, and passed through wind speed.

That "gust" you and hyevpranglostoph keep referring to in the record run, was a direct result of an error created by the CV's acceleration. If you look at the data from all the runs, you will find the same "gust" of wind while the CV accelerates every... single... time. And you will find the same huge changes in wind direction every single time the CV accelerates or decelerates through wind speed.

You will also find the only times during runs that the wind vane passes through the null are at those points.

If the cart was just barely able to beat the wind speed, that would be a problem, but since it kicked the wind's butt, it isn't.


I was started when the wind was 50 degrees off ddw. It does speak of bias.
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No, it doesn't, it speaks of real world winds.

I have cleaned it up so all the noise is removed.
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You have done nothing of the sort.
 
fttw said:
I'd like to see profpangloss's work that (he claims) shows the cart can't go ddwftw as claimed.

I'm happy for him to use a frame in which the earth is and remains stationary regardless of the motion of the cart; that will be true in the limit of the earth having infinite mass and there's nothing in the theory of the cart that demands the earth be finite for it to work.
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Here are a couple of posts containing such work:

http://www.internationalskeptics.com/forums/showpost.php?p=7162161&postcount=5181

http://www.internationalskeptics.com/forums/showpost.php?p=7176246&postcount=5631

Enjoy!

(The mistake in the second post is easy to state: He incorrectly invokes conservation of momentum for a system subject to external force: the cart, which is part of his system, is subject to a force from the earth, which is outside his system. The first post is harder to analyze. I believe, tentatively, that the problems with it can be summarized by saying that he considers frictional losses to be much greater than they actually are, perhaps by counting some of them more than once.)
 
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Perhaps, as in the past, Harold's shame at his egregious errors will make him disappear for a while. It happened before. Both humber and Harold disappeared for a while after the Blackbird had its irrefutable successes. But the stupid built up in humber until he couldn't help himself and he posted again. A couple of months later Harold came back too. But here he has made mistakes in physics that a high school student wouldn't have. When he sobers up he hopefully will be gone for a while. Too bad we can't say the same about humber.
 
Llyricist said:
LOL, the wind vane stalled for 20 seconds in an 8 to 20 MPH wind? You are too funny.
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You have no idea what vane stall is.
To claim that as ddw is funny, not only to me, but anyone who is not a cult member.

"ddw in 100 degree changing wind". LOL

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When a car is accelerating or moving against air resistance, the energy to do so comes from the gasoline burning in the engine, not from the Earth.

But how is it that the Earth does no work on the car, though it exerts a force on it as it moves? What happened to 'work equals force times distance'?

I believe the answer has to do with the fact that the Earth exerts a force only on the part of the tire in contact with the ground, and this part of the tire is momentarily motionless. So, while there is a force, there is no distance.

Is this essentially correct?

And can anyone expand on it?

Thanks.
 
I believe the answer has to do with the fact that the Earth exerts a force only on the part of the tire in contact with the ground, and this part of the tire is momentarily motionless. So, while there is a force, there is no distance.

Is this essentially correct?
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Yes.

However a real tire slips a bit, which generates a frictional (dissipative) force.
 
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69dodge said:
When a car is accelerating or moving against air resistance, the energy to do so comes from the gasoline burning in the engine, not from the Earth.

But how is it that the Earth does no work on the car, though it exerts a force on it as it moves? What happened to 'work equals force times distance'?

I believe the answer has to do with the fact that the Earth exerts a force only on the part of the tire in contact with the ground, and this part of the tire is momentarily motionless. So, while there is a force, there is no distance.

Is this essentially correct?

And can anyone expand on it?

Thanks.
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the work being done is considered at the axle, bearings and whees in a ground frame analysis. And has to be considered in terms of rotational KE, rather than linear. Of course, the results are the same either way, so using the linear calcs still come up with the same numbers.
 
Llyricist said:
the work being done is considered at the axle, bearings and whees in a ground frame analysis. And has to be considered in terms of rotational KE, rather than linear. Of course, the results are the same either way, so using the linear calcs still come up with the same numbers.
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I'm not sure what you mean about rotational kinetic energy. I'm talking about linearly accelerating the body of the car (or moving it against air resistance). Pretend the wheels are massless.

When you say that the work is done at the axle, bearing, and wheels, what does work on what?
 
69dodge said:
I'm not sure what you mean about rotational kinetic energy. I'm talking about linearly accelerating the body of the car (or moving it against air resistance). Pretend the wheels are massless.

When you say that the work is done at the axle, bearing, and wheels, what does work on what?
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That gets too friggin complicated for me to explain I'm afraid, which is why I'm happy to use the linear substitute. :)
 
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Llyricist said:
That gets too friggin complicated for me I'm afraid, which is why I'm happy to use the linear substitute. :)
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What do you mean by "linear substitute"?

I can understand how the expanding burning gasoline provides the energy to push the car forward if I replace the whole engine and transmission and everything by a horizontal cylinder attached to the car, with a piston in it that pushes back against, say, a brick wall when the gasoline in the cylinder is ignited. The gas pushes back against the piston and forward against the front wall of the cylinder. No work is done on the piston because it doesn't move, but work is done on the cylinder and attached car because they move forward.

Is that roughly what you had in mind?
 
69dodge said:
When a car is accelerating or moving against air resistance, the energy to do so comes from the gasoline burning in the engine, not from the Earth.
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Correct.

69dodge said:
But how is it that the Earth does no work on the car, though it exerts a force on it as it moves? What happened to 'work equals force times distance'?
I believe the answer has to do with the fact that the Earth exerts a force only on the part of the tire in contact with the ground, and this part of the tire is momentarily motionless. So, while there is a force, there is no distance.
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You can move a vehicle, by using a pole, as boatmen once did along the canal wall. There is no work done by the ground in that case.
A tyre is a similar, but continuous, version of that pole.
( There is a clue as to why the TM must be false)

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The rolling motion of the tyre is not like sliding friction. If it were, wheels would be of little benefit.
A tyre has a flat contact patch, which must be non-zero if there is to be rolling: if there were no friction, then the wheel would slide, but not roll.
As the tyre turns, the radius at the contact point falls, so it is quite impossible for the the wheel to make one turn, and have the vehicle move exactly the circumference, nor with out flexing of the tyre as it squeezes through the smaller cross-sectional area of the contact patch.

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(Another clue about the TM.)

The resulting rolling resistance, is the force and energy to do that work on the tyre, and the small "slip" with the ground as it does so. Otherwise, there is no work associated with the force applied to the ground.

The amount of work does depend on f*d, but is not simple.
1) The tire is locked up, and skids a distance X.
2) The tyre is rolling, not skidding at all. When idealised X =0
3) The intermediate case, is a mixture of 1 and 2 and changes with road surface and acceleration. That is a practical wheel and tyre.

In the case of pure skidding, all the work is dissipated at the tire/road interface. In the case of pure rolling, none of it is dissipated at the tire/road interface. In the intermediate case, you have no idea how much dissipation there is, unless you know X, the amount of slip.
 

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69dodge said:
What do you mean by "linear substitute"?

I can understand how the expanding burning gasoline provides the energy to push the car forward if I replace the whole engine and transmission and everything by a horizontal cylinder attached to the car, with a piston in it that pushes back against, say, a brick wall when the gasoline in the cylinder is ignited. The gas pushes back against the piston and forward against the front wall of the cylinder. No work is done on the piston because it doesn't move, but work is done on the cylinder and attached car because they move forward.

Is that roughly what you had in mind?
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I've got too much blood alcohol content at the moment to do an explanation justice.
 
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