Wikipedia:Reference desk/Archives/Mathematics/2013 July 19

= July 19 =

hf unit in tig welding
What is application of hf unit in tig welding. — Preceding unsigned comment added by 182.19.18.17 (talk) 09:41, 19 July 2013 (UTC)


 * Our article on TIG welding says "To strike the welding arc, a high frequency generator (similar to a Tesla coil) provides an electric spark; this spark is a conductive path for the welding current through the shielding gas and allows the arc to be initiated while the electrode and the workpiece are separated". Gandalf61 (talk) 10:03, 19 July 2013 (UTC)

Two brass crosses
A major gift to my church several years ago was two brass crosses, each of which sits on a base (we'll call it a supporting structure) about as large as the cross itself. It appears the front and back of each supporting structure is an isosceles trapezoid and the top and base of each supporting structure is a rectangle. The sides of each base are trapezoids but not isosceles. I shouldn't let my mind wander while in church but I was thinking there must be a formula for figuring out what the difference would be in volume if the sides of the bases were isosceles like the front and back. In other words, let's say there's an angle theta between the base and the front. If this were cheese, one could take a knife and cut a straight line from the top to the bottom such that when the base is reached, the angle would also be theta. Then there would be cheese left over with the added volume.

Let's say the top of the supporting structure has sides of length s1 and s2. The base has sides of length s3 and s4. The ratio of s4 to s2 is greater than the ratio of s3 to s1, though after the experiment with the cheese, the ratio of the now shorter side s5 to s2 would be equal to the ratio of s3 to s1.— Vchimpanzee  ·  talk  ·  contributions  · 21:02, 19 July 2013 (UTC)


 * The volume is the "average of top and base areas" multiplied by the height, so the volume will be the same whatever the angles (provided that the height remains the same). The volume is invariant under horizontal shearing.   D b f i r s   12:17, 20 July 2013 (UTC)


 * No, actually it won't, because the slice of cheese would have volume. Let me try. s4 is the actual length of the base. s5 is the length of the base after the cheese is sliced. If the new base is proportional to the top, s3/s1 = s5/s2. s5 = s2s3/s1. I don't know how to do equations here ... (s1s2 + s3s4)/2, multiplied by the height, is the actual volume. After the knife has sliced the cheese, you have (s1s2 + s3s3s2/s1)/2, multiplied by the height. I know that should be s3 squared.— Vchimpanzee  ·  talk  ·  contributions  · 16:05, 20 July 2013 (UTC)


 * Oh, I see: you are not shearing, but taking a slice off to make the sides isosceles. I think I over-generalised in my reply above.  My formula for volume works when two sides are rectangles, but not when all four are trapezoids. Apologies for misleading you.  The correct formula for the volume of your shape uses the Heronian mean, not the ordinary average.  If B is the area of the base and T is the area of the top, then the correct formula for volume is h/3(B + T + the square root of BT).  This makes your substitution more messy.  Sorry!    D b f i r s   19:36, 20 July 2013 (UTC)


 * I'll just give up on it. Interestingly, the sermon Sunday was about distraction, and I happened not to be sitting where I could see those crosses.— Vchimpanzee  ·  talk  ·  contributions  · 20:55, 22 July 2013 (UTC)