User talk:RainerBlome

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[edit] Pythgorean theorem & law of cosines

These are pretty firmly established math concepts for triangles, so the formula works. Yes, actually, really - as you have discovered: Merry Christmas! I see that you don't like having the Pythagorean theorem & Law of cosines mentioned in the article about spokes. Why do you consider your box (which I can't follow) to be better? Metarhyme 22:10, 25 December 2005 (UTC)

When I first found the formula on the web, I tried to understand why it should be correct. So I thought "How do I compute the length of a spoke?" and saw that a spoke is the diagonal of the described imaginary box. My explanation of the box is supposed to be understandable by everyone who knows basic trigonometry. Apparently it has to be improved. A picture would be nice here (fixed).

The Pythagorean theorem is implicitly used twice to compute the length of the boxes diagonal (d²=(a²+b²)+c²). This is true for all boxes, so I linked to rectangular box and saw no need to explicitly state that information on the spoke page. However, I just checked the box page and indeed, it does not show how to compute the length of the diagonal (fixed).

I do not see where the law of cosines should be applied here, can you explain? Which would be the sides of the triangle, where would the angle be? --RainerBlome 10:15, 3 January 2006 (UTC)

c^2 = a^2 + b^2 - 2ab \cos C . \;
looks like the law of cosines to me. The Pythagorean theorem is explicitly used once, in the first part of the spoke length calculation formula. The c in c² comes from the second part of the spoke length calculation formula - and that is the law of cosines, I would say. Do you agree?

Hm, "first part", "second part", what are you referring to? I do not understand your explanation. Anyway, I had a third look. Now I see how to apply the law of cosines to get the spoke length formula and added a corresponding paragraph to the article. Using the law of cosines yields (of course) the same result as doing (r_2 - r_1 \cos \alpha)^2 + (r_1 \sin \alpha)^2 = r_2^2 - 2 r_2 r_1 \cos \alpha + r_1^2 (\cos^2 \alpha + \sin^2 \alpha = 1). You might say that this proves instead of uses the law of cosines. Going this way has the advantage (for me) that it's more elementary, no need to remember the law of cosines.

No original work is permitted in wikipedia, so if you figure something out, you need to find where someone else did it (citation) before placing it in article space.

The link to my source is there. The derivation section makes it easier to verify the formula.

You found the sides and angle with your look number three. A 3D diagram would be clearer. If I make one, I'll show it to you before I post it. I may add a technique to get accurate dimensions for cases where ERD and hub measurements don't exist or are suspect - akin to the calc walk through - and post it to Talk if I get around to it. Metarhyme 20:55, 3 January 2006 (UTC)

[edit] Hi there!

Seriously, I'm making this edit to say hello because this is the loneliest talk page for an active contributor I've ever seen. Keep up your good work. Keegantalk 05:27, 15 July 2007 (UTC)

Thanks. You too! --RainerBlome 11:40, 15 July 2007 (UTC)

[edit] Moore-Penrose pseudoinverse

Jmath666, Let's continue the discussion on the article's Talk page, where it is easier to associate and salvage discussion results to the article. --RainerBlome 09:03, 17 September 2007 (UTC)