Page 1 of 1

### Cone Diagram beam width interpretation

Posted: Wed Nov 08, 2017 2:57 am
Hi All,

This may be a very simple question. From what I have read, beam width is indicated at the point where the light intensity is 50% less than that of the centre beam. For example, for a 60 degree symmetrical downlight, this would be 30 degrees either side from the centre beam. Viewing the Dialux cone diagram below, the beam width numbers are far less than 50%, what am I missing here?

### Re: Cone Diagram beam width interpretation

Posted: Wed Nov 08, 2017 8:58 am
Dear Stewart,

The illuminance cone is used to illustrate the beam patterns of rotation-symmetrical luminaires (often downlights or spots). It represents the light bundle emitted at one-half-peak divergence or half-value angle. This is the term used for the aperture angle at which maximum luminance is halved (cf. polar diagram). In general the maximum value is achieved at a beam angle of 0°. In this context, DIN 5037, Part 4 describes one-half-peak divergences of up to 5° as ultra-narrow beam, from 5° to 10° as narrow-beam, from 10° to 50° as wide angle and above 50° as extra-wide angle.
The illuminance cone diagram also shows mean illuminance available within an area delimited by the half-value angle (light cone diameter) for various mounting heights (cf. height above working plane). These data are used for approximatively lighting planning calculations. Image1.jpg (11.56 KiB) Viewed 3506 times

How to read it?

First we have to understand “how to make it”.

Look into the luminous intensity distribution, table or diagram. See Picture below.

Search the maximum intensity. In this example it is given at nadir (C0, Gamma0 or in IES wording H0, V0). Image2.jpg (33.13 KiB) Viewed 3506 times

The value is approximately 1100cd/klm.

Now we have to divide this value by 2 -> 1100 : 2 = 550

Now we have a look into the diagram and we check at which angle the luminous intensity comes down to 550cd/klm.
This is approximately the case at Gamma 10° (IES wording vertical 10°). Of course it is easier and more accurate to do in the table than in the diagram.

The value 550cd/klm is found to the left (C0 or H0 and to the right C180 or H180). In both directions it is 10° so together it is 20°.

The half-value angle (or in American English one-half-peak-spread) is approximately 20°.

If you want to have it correct, go into the table: Image3.jpg (15.21 KiB) Viewed 3506 times

I max is 1102cd/klm. Divided by 2 is 551cd/klm. With linear interpolation 551cd/klm is found at 8.9°. The one-half-peak-spread is 2 x 8.9° =17.8°.

And now how to read it??

The cone diagram makes it accurate and visible: Image4.jpg (35.55 KiB) Viewed 3506 times

This diagram shows the light source in the top as the source of the “light”. The yellow triangle is the “light” distributed in a one-half-spread of 17.8° (which is 2 times 8.9°)

On the left we can see the distance from the source in meter. 0.5, 1.0, 1.5,…

In the yellow triangle we can see the diameter of the circle created by the light in this distance. 0.16, 0.31, 0.47,…

On the right we can see the illuminance (E, lux) created by the light source directly below (C0,Gamma0 or nadir) given as “E(0°) and in the horizontal zero plane at the Imax/2 value “E(C0) 8.9°) in the distance as given on the left side.

For example:
In two meters distance from the spot, directly below I will have 83lux.
In two meters distance 0.315cm to any side (at the edge of the bright circle caused by the light) I will measure 40lx.

This tool is often used to find correct lights for spotlighting or correct downlights for corridor lighting. This is a fast and good way to preselect a product, before the calculation is done in DIALux.

Best regards,
DIAL Support Team