angles? (yes angles not angels)

[breezer]

I have a maths question, I cant figure out the answer, and cant think what to ask google.

So i will ask here.

Please note Its the calculation and how you get there i need to know.
and my drawing is not to scale

Its also one of my "silly projects"

The yellow box is a light, 1 foot across

The 20 is the total angle in degrees at this point

(10 degrees either side of centre)

1,2,3,4 at the side are feet

my question is this

at each foot away from the light how wide will the light beam be? (in feet)

I already know that at 0 feet away its 1 foot wide

its been bugging me all week end, and i am probably overlooking the obvious

 

[nelsy]

I think this could be your solution - sorry can't draw diagrams to help with this explanation.

Use a right angled triangle on each side (symmetrical). The central section is always 1 foot wide, the triangles on either side are growing in proportion to the distance away from the beam. The angle on each slant side is 10 degrees. Then use trigonometry to find the extra widths:

When 1 foot away, the extra beam on one side is 1 x tan 10 - therefore the beam at that distance is 1 + 2 x 1 x tan10 (central section and extra at both ends).

When 2 feet away the width of beam is 1 + 2 x 2 x tan10.

When 3 feet away the width of beam is 1 + 2 x 3 x tan10.

When 4 feet away the width of beam is 1 + 2 x 4 x tan10.

Therefore for a general distance of X feet away, the beam width is

1 + 2 x X x tan10.

 

[ninebob]

Struggling to remember how myself, but it can definitely be worked out, and I reckon it's something to do with this: http://www.gcseguide.co.uk/sin,_cos,_tan.htm

We know the "adjacent" side is one foot, we are trying to get to the length of the opposite. When we've got that (lets call it x) we know that the total width of beam will be 2x+1 at 1 foot away, widening by 2x for every further foot.

EDIT: nelsy got there first, looks like we're on the same track though!

 

[JohnD]

Non Angeli sed Angli?

 

[Softus]

1 + 2 x X x tan10.

Only if the sides of the beam are straight...

 

[breezer]

Thanks guys. too early in the morning for me to follow, i will look later. but thanks again

 

[tim west]

The beam oval width for any throw distance which i presume is what you are after is determined by the formula X=D*(2*TAN(F/2)) where:
X= Beam Oval Width
D=Throw distance in direct line
F=Beam or field angle (20 degrees in your case) so at
D=1 F=20 X=0.35
D=2 F=20 X=0.71
D=3 F=20 X=1.06
D=4 F=20 X=1.41

of course the beam oval length(Y) will change if the angle of throw to subject is not square you will need to know
The vertical distance from the subject V
The Focus angle off vertical K (angle =plumb bob vertical from source to line of source)
and the beam angle again F
The formula for this is
Y=V*(TAN(K+F/2)-TAN(K-F/2))

Hope that helps

 

[securespark]

Non Angeli sed Angli?

Reminds me of my old school's motto:

Non Sibi Sed Omnibus.

(or, if you're a pleb of 11, "No Silly S*ds On The Bus")

 

[empip]

Where 'd' is the distance from yellow box 'window' plane.
[2.d.tan(10°)] + 1

Tan (10) is approx 0.176 327

So at 4 ft ... [2 x 4 x 0.176 327] + 1 = 2.41' ish feet wide

 

[tim west]

so as to not cause confusion empip the beam angle is actually 20 degrees in the formula i supplied above, being the angle from beam edge to beam edge in that particular formula

 

[empip]

so as to not cause confusion empip the beam angle is actually 20 degrees in the formula i supplied above, being the angle from beam edge to beam edge in that particular formula

But did you anywhere consider the 1 foot width of the 'light'?

This is how I saw it ....

8.46 ins x 2 = 16.92 ins divide by 12 = 1.41 ft. plus 1 ft = 2.41 ft

 

[tim west]

so as to not cause confusion empip the beam angle is actually 20 degrees in the formula i supplied above, being the angle from beam edge to beam edge in that particular formula

But did you anywhere consider the 1 foot width of the 'light'?

This is how I saw it ....

8.46 ins x 2 = 16.92 ins divide by 12 = 1.41 ft. plus 1 ft = 2.41 ft

Not sure about your diagram or what you are referring to empip your diagram is suggesting that the total beam angle is 20 degrees yet the diagram shows two 10 degree angles with a missing piece of that angle inbetween suggesting that the 20 degrees quoted in your diagram is wrong?
The beam angle is still 20 degrees whichever way you look at it, the beam oval width is derived by using the first formula i described when lamp manufacturers specify beam angle it's the entire beam from edge to edge, the formula specifies this angle(total angle) to be used. the area the light hits will increase four fold over distance increasing by double(inverse square law), the intensity or energy drops accordingly to the law but the beam angle remains true over any distance otherwise the light would be bending (einstein theorists jump in at this point... but in the real world...)

 

[noseall]

very strong gravity can bend light. apparently.

 

[tim west]

Told you!

 

[empip]

Ok with it now !!

TW has the correct formula ..

Breezer's 1 foot wide description threw me.

If you look at my sketch the measure point would have to correctly start from the lower intersection of the beam edges and not at that section of beam being 12 ins wide, this section of the beam is at 34.0277 ins from the measure point 34.0277 x [2.tan10) = 12 ins
Add the 34.0277 or 2.835 ft to my 'distances' / TW's 'throw' and the thing comes right from origin using his formula 6.835 x (2.tan10) = 2.41 ft. wide.
It looks like the distance is measured from somewhere in or between the reflector and the front lense.