MrMongoose

Aren't ions in crystals of normal matter continuously vibrating anyway?

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I assumed he meant parallel, but as for the cycloid... Why? I'm not sure here, but from observing my trusty spring, it looks as if thats just the general case for the projection onto any plane, and just as the circle is one extreme, for this other extreme, we can specifically say its sinusoidal.

I understand why people are concerned, but it seems as ridiculous as not developing a chemical which could potentially kill cancer because it could potentially be used to kill Chinamen.

What has this got to do with militarisation? I didn't realise that it's an act of war to dissemble your own equipment.

It should just be a sinusoid. If you imagine a plane with the helix axis normal to it being moved along the helix, the intersection of the helix and the plane will move around in a circle at a constant rate. The distance of the intersection from the axis of the helix is constant, but the displacement expressed as as two orthogonla vector components, i.e. projections onto two orthogonal planes will be two orthogonal sinusoids. You can pick any plane and the projection will be a sinusoid with the phase depending on the plane chosen.

Table? Where?

It says image, not specimen. I'd say the obvious answer is that it shows things how they really are. Most people assume that a microscope shows things the right way round.

There will certainly will be coriolis forces- one from the rotation of the Earth about its own axis and another from the rotation of the Earth about the centre of its orbit, though they will be negligible compared to the other effects mentioned. The rotation of the earth will also have an effect on the climate, and wind patterns have an effect of surface soil formations which may cause slight gradients in certain directions. Though, these will cause gravitational components along the ground in predetermined directions rather than in the centripetal direction. Also, as the Earth spins and day becomes night, wolves come out and will chase you in a direction you hadn't planned on heading in!

Also, are you familiar with the word please?

This is a brain teaser?

I wish my job was that easy!

The point is that it isn't a 6" cylinder projected through a sphere and the overlap discarded. It is a 6 inch deep cylindrical hole in the sphere, so it MUST pass through the centre and its ends must be on the surface of the cylinder. What I think should have been added to the original question, however, is that the hole goes through the sphere from one surface to another. I just assumed that that was what was meant as there was apparently a unique solution.

Can't argue with the third one so I'll go about fixing it, and I agree the first thing is a bit shoddy. As for the second, I said they move closer in terms of angle, i.e. the cylinders length increases with respect to its radius. Again perhaps I should have been clearer! Might as well get your moneys worth;)

I think this is right... I take it that if the hole is 6 inches deep and goes all the way through, then there are a limited set of solutions- one for each sphere. As the sphere gets larger, the edges of he whole move nearer (in terms of angle not necessarily size) to each other. If we take a calculus based approach here, a sphere of radius r is made up of a series of infinitessimal disks, but when the cylindrical hole is drilled, it becomes a series of infinitessimal annuli. moving upwards from the "equator" with h increasing , the radius, y, of any one disk is given by [math]y=(r^2 -h^2)^\frac{1}{2} [/math] Taking a cylindrical hole of radius x out with the cross section of the cylinder to what I previously called the "equator", the width on an annulus w is, [math]w=(r^2 -h^2)^\frac{1}{2} -x [/math] And when the value of h is equal to 3 (i.e. the limit of a 6 inch hole, half being each side of the equator) the width of the annulus is 0, thus, [math]0=(r^2 -3^2)^\frac{1}{2} -x [/math] Which gives [math]x=(r^2 -9)^\frac{1}{2} [/math] And the volume of an annular element is thus given by [math]\frac{dV}{dh}=\pi((r^2 -h^2)^\frac{1}{2})^2-\pi((r^2 -9)^\frac{1}{2})^2 [/math] [math]\frac{dV}{dh}=\pi(r^2 -h^2-r^2 +9) [/math] [math]\frac{dV}{dh}=\pi(9 -h^2) [/math] Summing for the hole depth of 6 inches, [math]V=\pi\int_{-3}^{3}(9 -h^2)dh [/math] [math]V=\pi[9h-\frac{h^3}{3}]^3_-3[/math] [math]V=\pi(27-9)-\pi(-27+9)=36\pi inches^3[/math]