David Ashton
Well-Known Member
Steve, could we have some examples of 2.4k dimmers which only take 1.9k as I've never come across any in Australia?.
I prefer Mike Wood's formulae, detailed here: http://www.controlbooth.com/forums/collaborative-articles/7664-mathematical-formulas-lighting.html....Someone wanting brownie points can go back into the archives and find Ship's percentages for lamp life change, colour temperature change, output change etc with over or under voltaging a lamp.
Steve, could we have some examples of 2.4k dimmers which only take 1.9k as I've never come across any in Australia?.
Anyone want to show Pip where he went wrong?...Consider the almighty power equation:
Watts equals Volts times Amps:
W = V * A
You have 4 instruments, 575W each, for a total of 2,300W (2.3kW):
4 * 575W = 2300 W
That's cutting it fairly close if you're on a 2.4kW dimmer. Let's look at that in terms of current (Amperage)
Going back to our power equation...
2300W = 120V * A
A = 2300 / 120 = 19.17A
Anyone want to show Pip where he went wrong?
Again the follow-up question:
Given four fixtures with HPL575/115 lamps; which of the following currents would flow through the 120V dimmer?
a) 19.2A
b) 20.0A
c) 20.8A
d) 24.8A
Crap... ^^ I checked my math... O noes! What did I do? *shakes head*
Some dimmers with thermal breakers can only be loaded to 80%, or 1920W on a 2400W dimmer.
As I said in an earlier post Pip...reread the original question. There's a very important piece to this puzzle you're walking right over.
Here's a hint, your answer to Derek's question would be A).
And for the record it would be wrong.
Also for the record. 4 HPL 575w 115v lamps on a 20a 2400w dimmer is a bad idea.
(((psssst Pip! all the information you need is in the previous sentence)))
Said by STEVETERRY himself
Unless it's an ETC Sensor
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