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Maths help needed…

before i get a pencil and paper and start scribbling, you won’t just be tilting the light if the centre of the ellipse remains at 30m: you would actually have to rotate the light ‘over and down’ ever so slightly.
But i guess this is a theoretical rather then practical stage lighting question?
Posted 8 years agosimple trig isn’t it?
Posted 8 years agoooh good one. I used to like stuff like this when I was at school. Now I can’t remember – or never knew! – how to calculate this but to be honest I’d just cut some bits of card out and make a scale model of your beam, rotate it on the axis of the bulb until you get your required horizontal axis and then “upscale”.
The problem I see is that while simply tilting the light will give you your required 7.9m, the light will no longer be at the centre of that ellipse. Surely only an elliptical light beam will give you that.
Posted 8 years agoNot stage lighting, but yeah the light will need to be positioned as though its on an arc along the longest axis so will decrease in vertical height.
Can you get your pencil and paper now?!
Cheers ๐
Posted 8 years agoIs it 32.583deg (from the vertical)?
EDIT: Sorry, need to subtract 5.994deg from that, so should be 26.589deg
EDIT AGAIN: Sorry, I’m answering a different question…
Posted 8 years agom How did you calculate that?
Posted 8 years ago…and the centre of the ellpise won’t be the centre of the beam proper as it were. Imagine cutting a cone at a slant. The centre line of the cone will not be exactly half way between the 2 ‘ends’ of the cut.
try shining a torch at different angles on a wall in the dark and looking at the bright spot in the centre in relation to the edge of the beam.
Is it important that the light is 30m away from the centre of the ellipse. This makes a difference to the calculation. Although it might be easier that way…
Posted 8 years agoAm I having a stupid day ๐
Posted 8 years ago
Can someone explain to me how a circular spot can be tilted to create an elliptical beam yet still be in the centre of the ellipse? Or does it just need to be the centre of the short axis and not the long?Can someone explain to me how a circular spot can be tilted to create an elliptical beam yet still be in the centre of the ellipse? Or does it just need to be the centre of the short axis and not the long?
Now I’ve reread the original question…
The centre of the ellipse can be a fixed distance from the light source, however this won’t be the centre of the beam.
Posted 8 years agoit wont be a regualr elipse, you’re cutting a cross section through a cone so it will be rounder (and brighter) at the “short” end.
if you want a perfcer elipse you’ll need to wall mount the spot and relect it from straight above the centre point
Posted 8 years agoYou need to work out the frustum of a cone. Haven’t got time to do it now, but the basics are here http://en.wikipedia.org/wiki/Frustrum and a bit of goalseek magic in Excel will see you right… probably ๐
Posted 8 years agooh
it would be easier to send the beam through an outline (think stencil) of te elipse and then set your height
Posted 8 years agoJulianwilson, the 30 metre distance is important.
It needs to be 30 metres distance from the horizontal axis that remains 6.3 metres if that makes sense (this is critical). If it shifts slightly in the vertical axis, I can live with that.
Posted 8 years agoIsn’t it just (more or less) the angle whose cosine is 6.3/7.9 ??
Posted 8 years agossimon – a stencil would mean loss of light output and I cant afford to attenuate the output at all ๐
Posted 8 years agoAre we all intrigued by what this is for or is it just me?
Posted 8 years agoNot looked at the wiki link but I think you need to work out the beam angle, relate the other two angles of a traingle made up of the beam and the floor and use the cosine or sine rule to calculate it.
Posted 8 years agoYou probably need to look at something like this
But as I’ve not done this sort of Maths since my CSYS in the early nineties, that’s about as far as I can help you.
Posted 8 years agoGiven that the masses on STW can answer anything….
OK….I have a theoretical light positioned shining vertically downwards at a height of 30 metres and the beam angle creates a spot diameter of 6.3 metres.
I need to create an elliptical spot, maintaining 6.3m along the shortest axis and a the longer axis needs to be 7.9m. The distance from the light to the centre of the ellipse needs to remain at 30m.
What’s the formula for working out the angle at which the light should be tilted to create this ellipse?
I’m knackered and now my head hurts…
Enlighten me?
Posted 8 years agoOK, centre line of beam should be 37.114deg from the vertical.
The long axis of the ellipse will start 14.351m horizontally from a point directly underneath the lamp.
The long axis ellipse will finish 22.251m horizontally from the same point.
The centre point of that axis will be 30m from the lamp.
Posted 8 years agoLike this:
Posted 8 years agoCheers m ๐
So the hyp on the elliptical diagram is 30 metres (centre line of the beam). I can’t see how you calculated the lengths of the two other sides (23.7m and 18.3m) as you have no values for the angles other than the 5.994 half beam to get the longest axis when the light is vertical.
[Sorry if it’s obvious]
Posted 8 years agoShould the radius of the spot on the lefthand diagram not be 3.15m?
Posted 8 years agoShould the radius of the spot on the lefthand diagram not be 3.15m?
Yes, well spotted (now fixed).
So the hyp on the elliptical diagram is 30 metres (centre line of the beam). I can’t see how you calculated the lengths of the two other sides (23.7m and 18.3m) as you have no values for the angles other than the 5.994 half beam to get the longest axis when the light is vertical.
For the above solution the centre line of the beam isn’t the middle of the ellipse. Instead, the lamp is 30m from the centre point of the long axis of the ellipse. As lots of people have said above, the two aren’t the same.The centre line will actually fall to the right of the centre of the long axis of the ellipse (about 30cm further to the right).
I’d suggest you ask DrJ for his solution as it looks very elegant!
Mine was a bit brute force; simultaneous equations involving the height and angle between vertical and centreline of beam – one based on the triangle of the beam, the second on the rightangled triangle created by the 31.12deg angle.
Posted 8 years agoWhile not averse to a bit of maths, could you not just use a vertical spot with a gobo?
Posted 8 years ago“Left as an exercise for the reader” as they say in text books (b@stards!)
(In fact it was just an approximation, but not too far from your exact solution!!)
Posted 8 years agoThanks all.
Posted 8 years ago
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