How to work out a latern angle size

[stuartsl]

hello,

 

when they say a soucre four has an angle size of 90 degrees, how is this worked out? and how can I figure out which angle size I will need for my venue,

 

Thanks

 

Stuart

[JimWebber]

90 Degrees (In this case) means that the "Cone" of light coming out of it is at a 90^O angle.

 

To calculate the beam angle required is a simple method of trigonometry - based on how far the lantern is away from where it is lighting, and the size of beam required at the end of the throw.

 

+++EDIT+++

Obviously, you will need to take into account that as the beam gets wider, and the distance increases, the level of intensity gets less, so it's not necessarily a good idea to hang a 45^o lantern at double the distance.

 

HTH

 

Jim

[alex_hughes]

90 Degrees (In this case) means that the "Cone" of light coming out of it is at a 90^O angle.

 

To calculate the beam angle required is a simple method of trigonometry - based on how far the lantern is away from where it is lighting, and the size of beam required at the end of the throw.

 

HTH

 

Jim

 

 

 

Please note this comment is not sarcastic;

 

Is that Pythagoras theorem or something more complicated?

[Just Some Bloke]

The easy way is to download LD Calculator Lite (which is free) and just type in the figures. It will tell you exactly which beam angle you require. Can't go wrong! ;)

[smalljoshua]

IIRC it goes somthing like.

 

Beam Dia = Distance x (2 x tan (Beam angle / 2))

 

Josh

[Wuddy]

Also, lantern manufacturers have data sheets showing beam angle, beam diameter at distance and lux level at distance and the basic formulea for calculating these values appertaining to their products.

[JimWebber]

Not quite Pythagoras, but close.

 

Assume you want to light a 2M area. from 4M away.

 

This gives you an Isosceles(?) triangle of Base = 2 and height = 4

 

divide the base by 2, to give a right angled triangle of 1M Base by 4M High.

 

ie) The Adjacent is 4

The Opposite is 1

 

Using SOHCAHTOA , we know that the Tangent of the angle is Opposite / Adjacent

 

ie) 1/4 = 0.25

 

We now want the angle. The ArcTangent of 0.25 is (Digs out calculator) = 14.03^o.

 

Now, if you remember, we created a right-angled triangle, we want to go back to our Isosceles triangle.

 

Therefore 14.03 x 2 = 28.06^o

 

In this case, I would be looking for a profile of at least 30^o Beam width, and focussing it down as required.

 

Jim

 

+++EDIT+++ Must type Maths lessons faster!

+++EDIT AGAIN+++ For correct spelling of Isosceles.

[Wuddy]

IIRC it goes somthing like.

 

Beam Dia = Distance x (2 x tan (Beam angle / 2))

 

Josh

 

and that simply is, {tan (beam angle)} x throw distance = beam diameter

[alex_hughes]

Awesome

[stuartsl]

90 Degrees (In this case) means that the "Cone" of light coming out of it is at a 90^O angle.

 

To calculate the beam angle required is a simple method of trigonometry - based on how far the lantern is away from where it is lighting, and the size of beam required at the end of the throw.

 

+++EDIT+++

Obviously, you will need to take into account that as the beam gets wider, and the distance increases, the level of intensity gets less, so it's not necessarily a good idea to hang a 45^o lantern at double the distance.

 

HTH

 

Jim

 

 

sorry so what is the cone of the latern?and where is the 90degree measurement taken from the throw of the light? is it taken from where the beam leaves the unit?

 

thanks

 

Stuart

[gareth]

Strictly speaking, it's from the focal point of the beam. But to all intents and purposes, yes, it's taken from the lantern.

[stuartsl]

Strictly speaking, it's from the focal point of the beam. But to all intents and purposes, yes, it's taken from the lantern.

 

 

so if I had a lighting bar say 5 metres from the stage and the point that I wanted to light, and I was using a source 4 with a beam angle of 50 degree what would the light spread be on the stage? and what the easy (understandable way) of workin this out,

sorry my maths is nooo good and the LD lite calculator just confuses me!

 

Thanks for your help

 

Stuart

[Wuddy]

Ok,

 

Using {tan(beam angle)} X throw distance = beam diameter

 

{tan (50)} x 5Mtr = beam dia.

 

tan 50 (from tables or calculator) is 1.192

 

1.192 x 5 = beam dia

 

5.96Mtr = beam dia

 

lets call it 6 Mtrs.

[stuartsl]

Ok,

 

Using {tan(beam angle)} X throw distance = beam diameter

 

{tan (50)} x 5Mtr = beam dia.

 

tan 50 (from tables or calculator) is 1.192

 

1.192 x 5 = beam dia

 

5.96Mtr = beam dia

 

lets call it 6 Mtrs.

 

 

thanks thats a big help

[Just Some Bloke]

so if I had a lighting bar say 5 metres from the stage and the point that I wanted to light, and I was using a source 4 with a beam angle of 50 degree what would the light spread be on the stage? and what the easy (understandable way) of workin this out,

sorry my maths is nooo good and the LD lite calculator just confuses me!

 

Thanks for your help

 

Stuart

When you say the distance is 5m do you mean along the ground or "as the crow flies" to the lantern (in other words on the hypotenuse)?

 

If you mean it's 5m direct (rather than along the ground, in which case we need to know how far up the lantern is) then let me talk you through how to use LD Calc Lite:

 

First click on the very first icon (picture of a lightbulb).

 

In the "calculate" box select "beam oval width" as this will give you the width of the beam on the stage, which is what you want to know.

 

In the "knowing" box, looking at the plan above, you know D (the distance between the actor and the lantern) and F (the beam angle) so select "D and F"

 

Under "D" select the distance (5.00) and under "F" select the focus angle (50 degrees).

 

Read out the answer under "X=": 4.66

 

 

If you meant it was 5m along the ground, then we need to know the height at which the lantern is rigged. Knowing that allows you to work out the hypotenuse (which LD Calc calls D) so you can feed the data in and read out your answer.