2D Mag61 runtime, is my math right?

737mech

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I am putting together a Mag61 and was trying to figure out the runtime. Can someone with more knowledge in this department than me double check my work?

2D mag using 6x Duraloops at 2000mah each = 12000mah total capacity.
According to LuxLuthors chart on the 5761 bulb at 7volt he measured 5.4amps.

so 12000mah/5400mah that comes to 2.22 hours or 133 minutes runtime. So if I subtract 80% form 133 minutes that gives me about 106 minutes of usable runtime. Does this sound about right? Are my calculations correct?

Thanks for any help, a math wiz I am not!
:D
 
2D mag using 6x Duraloops at 2000mah each = 2000mah total capacity.

Corrected version.

If they were in parallel you could add them up to a 1.2V 12000mah. In series each battery sees the full amp load.
 
ah!!! I knew it was to good to be true. So I am looking at more like 30 minutes of runtime then. Thanks for the help, I am always learning on this forum. I really should have paid more attention in electronics class. (and math for that matter!)
 
Take capacity(2000mah=2A) divide by load(5.4A)=.37(%) x by 60(minutes in a hour)=22 minutes of runtime.I know there are other factors(amp load,battery condition,etc.)but this gives you a ballpark figure.Assuming my math is right?:thinking:
 
I run this setup too, and this is the conclusion I've come to. You guys can correct me if I'm wrong.

Current draw @ 7.2 V, as measured by LuxLuthor, is indeed around 5.4 amp. Resistance in the stock switch is around 0.200-0.250 ohm. So voltage sag would be (5.4 amp)(0.225 ohm average) = 1.215 V.

Voltage at bulb would then be 7.2 - 1.215 = 5.99 V.

At ~6 volts, Lux says the bulb will draw 5 amps. My setup draws just under 5 amps on a fresh set of cells, so I'm guessing that my previous assumptions and calculations are somewhat close to correct, or I got lucky.

From Silverfox's testing, a AA Eneloop delivering 5 amps will have a capacity of around 1750 mah. So (1.75 amp*hours)/(5 amp draw) = 0.35 hours = 21 minutes runtime.

So with the stock switch and a 7.2 volt AA NiMH battery setup, the 5761 is driven close to spec, and you can expect ~765 lumens at the bulb for about 21 minutes.

Is this pretty close?
 
Take capacity(2000mah=2A) divide by load(5.4A)=.37(%) x by 60(minutes in a hour)=22 minutes of runtime.I know there are other factors(amp load,battery condition,etc.)but this gives you a ballpark figure.Assuming my math is right?:thinking:

The formula I have always used, you multiply by 48 instead of 60. This is the formula DM51 posted in a thread awhile back. In using that formula would equal out to around 17 minutes.
 
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You can't just throw arbitrary constants into our ill-informed speculation! ;)

Edit: FWIW... I usually just roughly calculate it like ANDREAS FERRARI did, and when I want to put a little more thought into it, I'll try to account for the resistance like Point Source did.
 
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737mech I usually just calculate it just like ANDREAS FERRARI does, but in reality it usually is a bit less. ANDREAS FERRARI covered some of the other factors. My freshly charged battery packs of 6AA to 2D Eneloops usually measure about 8.4 volts so at the start you will pull alot more current which will instaflash the 5761 lamp. The Philips 5761 is a fine bulb but it really needs to be driven at 7.2 volts or less. You did not specify if you were going to direct drive it but I wanted to pipe in, in case you were. If you are using regulation to drive your build another way to estimate and calculate run time would be to measure the amp draw at the tail cap with a good DMM. Then take the capacity of your battery pack and divide it with the amp draw of your build. Happy Mods!
 
No regulator planned. I was just going to direct drive it. I figured there was enough resistance in the tailcap and switch to prevent the insta flash. But that was just an assumption. Thanks for all the help guys.
 
My freshly charged battery packs of 6AA to 2D Eneloops usually measure about 8.4 volts so at the start you will pull alot more current which will instaflash the 5761 lamp. The Philips 5761 is a fine bulb but it really needs to be driven at 7.2 volts or less..........

No regulator planned. I was just going to direct drive it. I figured there was enough resistance in the tailcap and switch to prevent the insta flash. But that was just an assumption.

The simple equation I gave you is only for runtime.

In this specific case Fulgeo is right-your going to blow your bulb the first time you switch on your light!!!!!!!!!!!:poof::poof::poof:

That bulb(and I've blown a few)will not take more than 7.2v.

A fresh battery pack will blow it-and the next bulb you install.

It is a touchy bulb to use without some sort of regulation or resistance.

Currently I only use them in my stock MagCharger.A wonderful light that I recommend to anyone that needs a ton of throw but doesn't mind a 2.5 lb. flashlight. :grin2:
 
Couldn't tell you to be honest, that's just what I've always went by. I've often wondered what the 48 was suppose to be as well. :whistle:
LOL...

I say all of you are wrong, and 46.31415926535897932384626433832795028841971693993 is the proper constant. :nana:

Just tried some searching around to see where DM said that, but even with refining the search down to 27 results, it's too much to sift through at the moment. Try sending a quick PM, and he might be able to tell you where it is.

I'll take a stab at the explanation though. Perhaps 48 minutes at 1C discharge is where voltage sag starts to cause a visually noticeable drop in lumens and/or color temp in most incan setups, thereby giving a more realistic estimate of the expected runtime?
 
I suspect the reason to use 48 instead of 60 is to account for the lower-than-advertised capacity delivered at high current draws. 48 minutes / 60 minutes = 0.80.

0.80*2000 mah = 1600 mah... I think that's a little low. Using SilverFox's testing once again, a AA Eneloop will deliver ~1750 mah at a 5 amp draw. 1750 mah / 2000 mah = 0.875.

Therefore, I'd like to propose a new arbitrary correction factor: 0.875. :D

Do with it what you will... :aaa:

:popcorn:
 
well crap, there goes the plans for my Mag61 build. I just bought a new 2D host, an adapter for 6aa's, and some 5761's. The parts won't go to waste but its back to the drawing board for me. Ive already got a 2D ROP Low running off of 2x18650's. Maybe I will just have a new ROP hi running off of 6xAA's.:duh2:
 
No need to give up 737. Like I said, I run this exact setup without problems. With no resistance fixes done to your switch, you could probably get away with it too.

The problem Andreas and Fulgeo mentioned involves fresh cells hot off the charger. To avoid this, just charge your cells and LET THEM REST OVERNIGHT (just to be sure), and then have at it. If nothing else, like you said, you still have your ROP High to play with.

Since you already have all the parts, its definately worth a shot.
 
I suspect the reason to use 48 instead of 60 is to account for the lower-than-advertised capacity delivered at high current draws. 48 minutes / 60 minutes = 0.80.

I think you're on to something there, I wish I could remember what thread it was posted in, I've been looking for it but I can't seem to find it. :shakehead

Either way, it's only an estimate, I just use it as a ballpark figure.
 
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