Recent content by CarlAK

Well if the flow is through a (relatively) small opening, the instantaneous flow rate is dependent on the pressure head, size of opening, and shape of opening. V=Cd (2gh)^1/2 Q=VA So tank geometry doesn't come into play. Time to empty would be affected by tank size & geometry.

You can probably assume the ground temerature to be constant and unchanging for first estimate. See this tread for basic equations & suggestions: http://www.eng-tips.com/viewthread.cfm?qid=171100&page=1

I believe you can use orifice flow formula to approximate the static pressure necessary to achieve a given flow through an opening. A 3.5" diameter orifice still behaves similar to a smaller orifice, it may have a little different Cd (coefficient of discharge) than other sizes. As I recall...

Solve for T from the sum of forces in Y equation, since the only unknown is T. Now substitute T into the "x Sum" equation to solve for the horizontal force. The x direction equation should be: 0 = Fh - Tsin 37 That equation does not imply movement, it is that the horizontal component of the...

Well I just have to disagree about the second triangle being equilateral, isn't possible if those dimensions are distances to the centroid. Every triangle has a perpendicular distance to the centroid of 1/3 the perpendicular height, for each side. So I still contend that both triangles have the...

OK in looking closer I see that I used a wrong assumption, that an altitude passes through the center of the triangle. That's only for an equilateral triangle. But, it is true that the distance from a center to a side (perpendicular) is one-third the altitude. So, you can still conclude both...

If you draw it with all the known information I think the solution will come to you.. Draw all three altitudes, which then forms 6 smaller triangles. you know the length of all 3 altitudes, as well as the distance between the vertex & the center of each (which is 2/3 of the altitude). Using the...

Actually I drew it in a CAD program. Given a "center" point 665 above a base (side); draw circles centered on that point of radius 2* 664 & 2*666 to intersect the base at the vertices, draw 2 altitudes of 3*664 & 3*666 from vertices through center point, then can draw the 2 sides.

The center of a triangle is 1/3 the distance along each altitude. So since the center of each triangle is 665 from the base, the vertical(?) altitude of each is 3 times that, 1995. But the second triangle is not equilateral. It has sides of 2303.6, 2307.1, and 2300.2 First triangle has all...

https://www.physicsforums.com/showthread.php?t=338738

Measure the length along the curve (beam), the chord (straight line distance from beginning to end of the curve, and the maximum distance between the straight line and the curve. Actually you only need any 2 of these measurements. Now draw a sketch of an arc, center of arc, radius, chord, angle...

maybe this time someone will "solve" this :) https://www.physicsforums.com/archive/index.php/t-11889.html

Guesstimate - 100 psf per floor, which includes dead + live loads.

Yes; as I recall the radiation heat transfer is a function of (T14-T24) and the radiation coefficient (don't recall the term), 1.0 for black body.

Yes, it may have had an initial higher temperature :) There may be wind, the black is more exposed, so warms faster There is other radiation besides the sun; the 2 radiate to each other, and black is more affected (I suppose this would tend toward equilibrium) Radiation from the sky-if the...