7x solar - 50.000 LUX/hour for 3 hours = 150.000 Lux on the solar app of the watch?

Yes. That´s how i understand it. Lets see if anyone can confirm as well :)

Gosem said:

50.000 Lux for 3 hours is this equal if the solar app shows 150.000 Lux 

Yes.  50k Lux-hours for 3 hours is the same as 150k Lux-hours for 1 hour of light imparted on the solar panel.  The real question: Is the watch capable of converting the 150k Lux-hours for 1 hour vs the 50k Lux-hours over 3 hours into the same amount of usable energy?  Probably.

Can confirm, everyone in here is correct. The formula is:

1 lux * 1 hour = 1 lux hour

So,

50k lux * 3 hours = 150k lux hours

Also,

150k lux * 1 hour = 150k lux hours

25k lux * 6 hours = 150k lux hours

10k lux * 15 hours = 150k lux hours

Actually no...

There was an information somewhere on the Garmin site that specifically said that it need 50k lux to achieve maximum power efficiency... so considering a best case scenario for solar panels of efficiency rating of around 20% - this would be only for the outer ring - for the glass itself - probably around 5%....  so I would say that anything above 50k would not be detected by the watch. The little icon on the solar widget shows the intensity on it as a 10 5k lux boxes  - (lumen per sqr meter). So if its is full you have over 50k luxes on the watch surface, but the watch will only use 50 for power conversion (it's the sad truth of the low efficiency of solar panels in our age and the size of the watches face)....

So the actual calculations will be based by the number of boxes on the widget ( it updates on around a minute or so)..:

1box = 5k lux

1box by 1 hour = 5k lux 

50k/1box lux we need for max efficiency = 50000/5000 = 10h for max efficiency 

50k/2box = 50000/10000 = 5h...

50k/15k(3box) = 3.33h 

50k/25k (5 boxes) = 2h

....

 10box = 50 000 = 50 000/50 000 = 1h

but this is to achieve maximum efficiency- not that it will charge it.... there is no information that I have found on the power capabilities of the solar sapphire or glass (how much it can give) and all depends on the battery draw 

in best case in normal non activity(with only hr) I think the maximum power that it can give is to sustain it without drawing of the battery - anything out of that will also draw current from the battery to fill the power gap. So the moment your out of those 50k conditions it will draw of the battery and in 0lx conditions - 100% from the battery..

So if watch needs 1 mili amp of current (just for visualization...) and max glass can give is 0.3 - the other 0.7 will come from the battery to have it 1 miliamp. If the watch needs 0.5 and the glass gives 1 miliamp then 1-0.5 = 0.5 milliamp will go to battery...(in the real world will be less due to power losses)...

If we know the capacity of the battery and we charge it to 100% with sun and it is off during that time we can guess the power capabilities of the glass and the power consumption can be extracted from that... but that's a lot of work and patience....

so for instance we have 1mah battery(that means that we can draw 1 miliamp current for 1 hour or 0.5ma for 2h....) and if we leave the watch on empty battery and turned off in 50k lux conditions for 1 hour and we have 50% battery(so 0.5 mah) then we know that the max that glass can give is 0.5ma. And then the same experiment with turned on watch = we can calculate the power draw...

But in the end Garmin engineers already know all the numbers and have done the calculations. They just need to release  the information (but I guess its corporate secret...) ....

Here is a better formula...

fenix 7x solar + sunlight = charging.

Unfortunately again no... It's  = more slowly discharge...

Ivaylo Minkov said:

so I would say that anything above 50k would not be detected by the watch

Yes, the solar panel design will set a peak power point, and as you point out, is probably set to the 50k solar input.  So simply extrapolating the output power curve as a linear function of increasing Irradiance would not hold (my simple math from above).  Certainly the panels still produce power above 50k lux and one can only imagine Garmin is not throwing away solar panel output above the peak power point.  I assume that the solar graphic is only feedback to the user that the watch is receiving the 50k lux solar input and not an internal electrical cutoff point.

Ivaylo Minkov said:

But in the end Garmin engineers already know all the numbers and have done the calculations. They just need to release

Not going to happen.  From a user's standpoint, does it really matter?

What would be more interesting for the user is a general +/- graph similar to an electric vehicle (pwr consumption vs. regen).  The +/- may not necessary be a direct mapping to absolute charge/discharge, but some sort of normalized charge/discharge rate.

LOL ok.

My point was to joke about the fact that people don't need to use all these formula's and charts to know that if they use their device in sunlight, the charge will last longer.