Recent content by Scott S

Ah! Nice! Thank you very much.

I'm not very familiar with statistics, so that's my main problem. I use a least squares software to generate the coordinates of points and their associated 95% error ellipses. If the error ellipses meet a certain pre determined tolerance then I'm done. However, I have a situation where I have...

OK, so a LONG time ago I believe I saw a small drawing by Newton which represented the first true drawing of a Cartesian graph with an x and y-axis and "O" at the origin. Now (that I'm old and can't remember what I had for breakfast) I'd like to find it but can't remember where it was. Any help...

Yes! Thank you.

Sorry, 30 years since college and I wasn't awake all the time. I'm trying to translate the symbology in the attached pics. Some of the super/subscripts are throwing me off in formulae 13, 14 and 15. Y and X = planar coordinates of points. n = number of points. σβ = standard error of angle...

Is there a relatively simple algorithm to compute the area in percentage under the curve as represented by a sigma value? For example; 3 sigma = 99.7 2 sigma = 95 1 sigma = 68.3 Now suppose I wanted to know 2.5 sigma without a table.

Can anyone point me towards a listing of common star names for the FK5? I've been going through Wikipedia, one star at a time compiling common name, constellation and greek letter for the FK5 and I have lost my enthusiasm after a hundred or so stars. Surely someone (with more fortitude)...

Got it, thanks. Wrong formulae.

Right, sorry. Right Ascension and Declination are spherical coordinates for stars etc., where declination equals latitude and right ascension longitude. http://en.wikipedia.org/wiki/Celestial_coordinate_system

OK, I'm really rusty on this. I need to convert a unit vector to right ascension and declination. I believe I recall the formulae correctly, as I seem to have gotten 1 as my radius. So, that's good. In the pic I have my unit vector (P2), my actual answer below, my expected answer in...

Woohoo! Thanks gneill, with a little tweaking of the values I managed to get the answer in INCHES and then converted to feet and looks like I'm golden. Thanks again.

g = 9.81m = 32.17405 ft x = 999.126m = 3277.972 ft A = 10.2mm^2 = 0.003 inches (from manufacturer - I get 0.034 feet) E = 2x10^5 N/mm^2 = 29,700 KSI (from manufacturer of tape) M = 0 = 0 (theoretically any attached weight kgs in orig/lbs...

gneill, That is it in a nutshell. I took a few tries at it and cannot reproduce the metric result with "english" units. I added a jpeg of the text explanation if that helps.

I have attached a pic which really explains the formula and context of the problem. I'm guessing a gravitational effect passes for physics... I am trying to convert the formula in (b) below into feet. If I could reproduce the answer with English units, I could move on and apply the...

I have attached a pic which really explains the formula and context of the problem. I am trying to convert the formula in (b) below into feet. If I could reproduce the answer with English units, I could move on and apply the formula to other situations, but the multiplicity of English...