hobz

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I am under the impression that a wind turbine generates lift because the rotor blades move through the air. At the blade tips, the speed by which the rotor blades move through the air, is much greater than the wind speed.
However, I fail to see how the incoming wind adds to the movement of the rotor blades?
Also; can wind turbines start by them selves, or do they need a certain jolt of energy (sort of like a starter on a gasoline engine) to begin converting wind power into lift?
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Mr Skeptic's idea led me to a solution. I finally figured it out. Both geometrically and trigonometrically.
The geo. way was based on the area of a parallelogram spanned by two vectors, and the trig. way was based on the difference formula for cosine.
@ moth:
Thank you for posting your notes. I've seen that lecture before. He does a very good job.
What program have you drawn your notes in?
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Similarly:
A car battery is not an ideal source, and cannot maintain a voltage of 12 V (or whatever the rating is) independent of the current drained.
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Yes. I neglected to inform that my starting point is the geometric interpretation, which can be understood knowing only trigonometry and very basic vector stuff.
From the geometric interpretation I would like to arrive at the standard definition.
I have drawn a figure which shows my current progress. I can upload it when I get a chance to scan it. If it helps, I can upload it.
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Interesting. I'll give it a go.
However, I would like the [math]a_1b_1+a_2b_2[/math] to "magically" pop out, and rather not employ a property of the scalar product, before I have proved the equality of [math]\vec{a}\cdot \vec{b} = a_1b_1+a_2b_2[/math].
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Yes, but it doesn't really help me express the area of a parallelogram in terms of coordinates. Besides, using the cross product to prove some property of the dot product, does not offer any insights as to why the relationship between the two sides, in my first eq., exists.
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I am trying to prove
[math]
\vec{a}\cdot \vec{b} = \vec{a} \vec{b} cos \theta = a_1b_1+a_2b_2
[/math]
in two dimensions.
I have come to the conclusion that I need to express the area of a parallelogram spanned by the two vectors
[math]
\vec{a} = [a_2; a_1]
[/math]
and
[math]
\vec{b} = [b_1; b_2]
[/math]
by their coordinates.
So far I have tried to express the height of the parallelogram in terms of these coordinates, but I have not succeeded.
Can you help me further?
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Thanks for answering.
So the printf function (or similar in other languages) converts the binary representations into strings  sort of an optical illusion.
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Since computers use binary numbers, how is it possible for them to display decimals?
For instance, the following (C code) will print out 10, and not a binary representation of 10.
int i = 10;
printf(i);
How is that done?
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Does your time criterium imply that quantum tunnelling can happen over infinite distances of potential barriers, as long as we "wait" long enough?
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The loss in potential energy of the particle.
Emitted as photons?
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Ahh, but of course. Thanks for spelling it out.
I have apparently developed a nasty habbit of ignoring constants.
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I insert the limits on the right hand side:
[math]
\frac{ba}{2} \int_{1}^1 f \left( \frac{ba}{2} x + \frac{b+a}{2} \right) dx
[/math]
[math]
\frac{ba}{2} \left( F \left( \frac{ba}{2} \cdot 1 + \frac{b+a}{2} \right)  F \left( \frac{ba}{2} \cdot 1 + \frac{b+a}{2} \right) \right)
[/math]
[math]
\frac{ba}{2} \left( F \left( \frac{b}{2}\frac{a}{2} + \frac{a}{2} + \frac{b} {2} \right)  F \left( \frac{b}{2}+\frac{a}{2} + \frac{a}{2} + \frac{b} {2} \right) \right)
[/math]
giving
[math]
\frac{ba}{2} \left( F(b)  F(a) \right)
[/math]
which is not equal to the left hand side.
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I fail too understand what properties of Nature are responsible for getting particles (representing energy) out of a local energy state minimum to an absolute energy state minimum.
Perhaps it is not fair to demand an explanation of the ways of Nature. In that case, what logic explains the act of tunnelling through a potential barrier without having the energy to do so.
Are the responsable mechanics related to the spontaneous nature of atom emissions and absorbsions?
Where does the lost energy go?
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Thank you for answering.
I see your point, and understand that then constant is there to "kill" [math]du = \frac{2}{ba}[/math].
However, I fail to see why substitution is needed. What is wrong with the logic I use?
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I was wondering why/how the following statement is true.
[math]
\int_a^b f(x)\quad dx = \frac{ba}{2} \int_{1}^1 f \left( \frac{ba}{2} x + \frac{a+b}{2} \right)
[/math]
I have come so far as to insert 1 and 1 instead of the x, and subtracting lower limit from upper limit. But that gives me
[math]
\frac{ba}{2} F(b)  F(a)
[/math]
which has a strange constant in front of it.
I have tried to interpret this constant as some sort of a meanvalue of interval. I don't know why it's there or what is means.
Can somebody shed some light on this matter? Thanks.
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As far as software for graphs goes, I'm trying to keep up a list here: http://www.freewebs.com/mytestingzone/Reviews/graphmath.xml
Looks interesting. Been looking for software for sketching math concepts for years.
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Walter Lewin's lectures on physics(!) are my personal favorites.
You can find them at MIT's open course ware.
8.02 Electricity and Magnetism
I would also recommend The Mechanical Universe And Beyond which is available for download if you are a resident of the US or Canada. It's a rather old series, but explains most physics in a way that requires little or no prerequisites, and has very good animations for most explanations.
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I will recommend principles of chemical science. While not quantum mechanics per se, it has a very good introduction to QM as well as the historical perspective of the theories and ideas leading to QM.
The level is undergraduate, and generally easier than Susskind's lectures.
Though many people find it hard to abstract from the teacher's very distinct voice, the ability to do so will pay off. Her material is well disposed and conveyed in a good way.
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Thanks for picking up on this thread again.
Does that imply that at higher frequencies (than that of an ordinary oven) only some of the waves escape, while other are reflected since there aren't holes all over, and where there are no holes the required wavelength to escape is smaller than that of an incoming wave?
Earlier on there was an explanation involving opposing fields that canceled out outside the oven, while creating standing waves on the inside. Is the true or false?
The real problem is picturing why it is obvious or at least reasonable to understand how a wavelength is related to something spacial as a hole in a metal.
The property of wavelength must have a specific interpretation in space, right?
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It's not opposing fields, per se, that causes reflection. It's the inability (or difficulty, if you'll excuse the anthropomorphic nature of that) for the wave to exist in the material, so there's a boundary condition in effect. You get reflection because you have to in order to conserve energy.
I have rethought this answer and am wondering why higher frequencies CAN exists in the material. I assume that they can, because they will travel through the grid of the oven. Why is that?
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Thanks for clarifying. I will look into the methods proposed.
It would seem that what I am calculating is related to the area of an annulus http://en.wikipedia.org/wiki/Annulus_(mathematics)
I follow you explanation, D H, but I am having trouble with visualizing what is meant by this on the wikipedia article:
The width of the ring can be determined by using Pythagoras' formula for a rectangular triangle with side lengths [math]dx[/math] and [math]f'(x) \cdot dx[/math], which leads to [math]\sqrt{1+f'(x)^2}\,dx[/math]
The infinitesimal surface area of the circular ring thus is equal to [math]2\pi f(x)\cdot \sqrt{1+f'(x)^2}\,dx[/math]
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I want to use circle circumferences to find an area. I don't see how I lack dimensions in the question.
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In the shower this morning, I suddenly got the idea of calculating the surface area of a sphere as an integral of circumferences.
My approach is as follows:
The integral is [math] 2\cdot \int_0^r 2\pi t dt [/math] where the "2" in front accounts for the surface area of whole sphere (equivalent to having [math]2r[/math] as upper limit in the integral  by symmetry).
The result is then [math]2\cdot 2\pi [\frac{1}{2}t^2]_0^r[/math] giving [math]4\pi \frac{1}{2}r^2[/math] or [math]2\pi r^2[/math] which is wrong.
Where did I go wrong?
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Wind turbine questions
in Physics
Posted
Thanks for replies.
I have figured this out. The blades are angled in such a way, that the relative wind to the blade is moving faster on the upper side, than the lower side (upper side faces wind direction). By Bernoullis principle, this causes the pressure on the bottom side to exeed the pressure of the upper side, causing a lift force on the bottom side. A component of the lift is in the direction of rotation, thereby adding force to the rotor.