Recent content by frogtag

Cheers this is a good starting base for me.

Can someone tell me what the formulas are for the x,y & z co-ordinates of a planet on its elliptical orbit of the sun? I don't mind astronomy terms (I'm no expert on them but have a basic knowledge and will look up and learn any I don't know). It's not homework, just something that's doing...

Hi a bit random this question, but I'm trying to divide a sphere into many equal shapes (round-ness isn't important). So far I've come up with a dodecahedron, 12 pentagons. This isn't enough of a division for my needs. Can anyone give me the names of geometric shapes that have more divisions...

t = (B - A) / (x - A) ?

B - A ----- = x - A t (B - A) / t = x - A (3.0-2.0) / 0.5 = 2.5 - 2.0 1.0 / 0.5 = 0.5 0.5 = 0.5 That works...

I have got that far already, used... t = 0.5, A = 2.0 & B = 3.0 to prove it, x = tB + (1-t)A x = 0.5 x 3.0 + (1 - 0.5) x 2.0 x = 1.5 + (0.5) x 2.0 x = 1.5 + 1.0 x = 2.5 then expanded the brackets, x = tB + 1A - tA x = tB + A - tA x = 0.5 x 3.0 + 2.0 - 0.5 x 2.0 x = 1.5 + 2.0 -...

Quick clue ... is this going to be a quadratic?

Okay ... x = tB + (1-t)A x - tB = (1-t)A or x - tB = A(1-t) x - tB = 1A - tA x - tB = A - tA Not much so far...?

If it is a homework question its about 16 years late! No it's just been a very long time since my A-levels and having 2 separate t variables is throwing me. But I'll have a go and post my failure in a bit!

Can anyone help me rearrange this equation so that t is isolated please, it's driving me nuts. x = tB + (1 - t)A all I can seem to do so far is isolate one of the t's... 1... x - (1-t)A = tB 2... x - (1-t)A ---------- = t B

Can someone just check my maths to see if this is correct. Hopefully, this should give the coordinates for a point on a sphere? x = radius x cos(vert angle) x sin(hoz angle) y = radius x cos(hoz angle) z = radius x sin(vert angle) x sin(hoz angle)

Marvellous. Thank you kindly for your help Tiny-Tim, you're a star! :biggrin:

Because its a computer program I have to use timed intervals to check on the position of x to see where it is. It's like taking a film of a ball moving and checking each frame to see where the ball is. So x will be considered a staionary point. For the moment yes. The process I'm taking is...

This equation will be for a computer AI program. The idea I have is that the vector represents a characters line of sight. Point A in the person. Point B is where they can see to. Now I want to be able to test to see if another person crosses the vector/the line of sight. Now using the...

Too say I'm rusty is an understatement! Using x = txA + (1 - t)xB surely when t = 0, x = xB not x = xA? at t = 0 ... 0 * xA = 0 and this is canceled out, but (1 - 0) = 1 and 1 * xB = xB, so x = xB? at t = 1 ... 1 * xA = 1, but (1 - 1) = 0 and 0 * xB = 0, this part is canceled...