I found 2 formulas in this forum thanks to @russellcresser but there is some variance when I look at sites online like moongiant or starwalk.
(numberFormat("#.#",(50+50*sin(rad([MOON_PO]*360/28-90)))))
(numberFormat(“#.#”,(50+((sin(rad((((([MOON_PO])/28))*360)-90)))*50))))
I assume that this has to do with location and/or time of day - is this correct and if so is there some middle ground formula?
Thanks for any insights!
Hi @andi-sf . Those two formulas are the same . One is my attempt at grownup maths . I am not a mathematician . I do not claim it is faultless . I like to keep it simple and ignore errors for time of day etc . Here is a formula that by all accounts takes care of the wobble . I ported it from a Facer Formula . So again I can not say it is perfect . You will see that it gives 0 to 1 . So you can use it how you like .
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((([UTC_TS]/2551442844-0.228535)+0.01765*sin([UTC_TS]/378902233-1.828)+0.00583*sin([UTC_TS]/5022682843-3.179)-0.00383*sin([UTC_TS]/437445587-4.186)-0.00057*sin([UTC_TS]/218722742-2.084))%1)
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I will post an appropriate credit Later .
I just wanted something as a number that is more accurate than the 8 slot named Phases and smoother than a 28 Image change .
Hi @russellcresser - thanks for another option.
I tested these out and the 2 older formulas give 40.4 while the new one gives 0.78. I also checked online at calendarr.com which is closer to the older formulas with 42.61%.
This looks to be a hard nut to crack within WFS… 
The long formula gives 0 to 1 for moon age starting at new moon . It is up to you to use the formula acording to your purpose . So for example if you multiplied it by 29.53 you would get moon age in days etc .
The Sin Rad formula I have used is a conversion as the amount of light per Moon age is non Liner . I seems to prove ok against som formulas but not others . At the end of the day it is usually within 5% . I feel that is pretty good . If you need something more accurate there are a few Apps to compare it to .
Hi @russellcresser. Thank you so much for following up and the hint on how this formula can be used. Thanks also for always being so helpful and so generously sharing your knowledge and advice in this forum - it is so much appreciated!