Help with some maths please!

[Steve1976]

Ok I was always miserable at maths at school so I need a little help please.

I'm reading a book at the moment which mentions how to calculate your approximate limiting magnitude on a given night. You take your aperture in mm (in my case 127) and divide by 6 then multiply that number by itself so 448.7 for me. Then the book says 'take the approximate logarithm to base 2.512'.

Can someone explain to me what on earth that means or even just give me the answer?!

I'm meant to add this number to the faintest star I can see with the unaided eye to give the limiting magnitude...

[russ]

Is the 6 your dilated pupil size in mm?

And Is this to find the limiting magnitude of the telescope rather than your skies?

There is an online form you can use that does the calculating for you. Just input the mirror size, pupil size and approx sky limiting magnitude:

http://www.cruxis.com/scope/limitingmagnitude.htm

[Demonperformer]

approximately 11.37 (11.37^2.512=448.2, 11.38^2.512=449.8)

But if you add 6 that gives you 17.37!! Think the calc is wrong somewhere.

[Dave Smith]

I think you may have that the wrong way round DP.

I think it is approx 6.63 i.e 2.512^6.63=448.9

So if the visual limiting mag is 6 (very dark site) then with the 127mm scope it would be 12.6 which seems reasonable.

Mag 17.4 is not possible.

Dave

[Demonperformer]

Dave, yes you are right. Had just reached the same conclusion, but you beat me to it.

Apologies for the misleading answer, Steve

[Steve1976]

Thats great guys thanks a lot.

Just out of curiosity how do I do that log calculation on a scientific calculator?

[Dave Smith]

2.512^L = 448.7 where L is log to base 2.512 of 448.7

take logs of this equation (to any base e.g. log or ln) to give

L x log 2.512 = log 448.7

therefore L = log 448.7 divided by log 2.512

L = 6.63

Dave

[Steve1976]

Ah I think I got it now!

So for my 12x50 binos L = 4.60. In other words if on a given night my naked eye LM was 5, through the binos it would be roughly 9.6 (-2 for faint fuzzies)?

[Dave Smith]

Ah I think I got it now!

So for my 12x50 binos L = 4.60. In other words if on a given night my naked eye LM was 5, through the binos it would be roughly 9.6 (-2 for faint fuzzies)?

Yes, that's what I get

Dave