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ZEISS DTI thermal imaging cameras. For more discoveries at night, and during the day.

Help resolving potential issue with UV 8x42HD+ (1 Viewer)

One way to eliminate nearly all the variables except exit pupil is to make a 32mm aperture mask and place it in front of one side of your 8x42. Using one eye, switch back and forth between the two sides.
 
And then put the mask in front of the other barrel or turn the binocular upside down :)

Or do both, as the transmission could differ between barrels in an older binocular, as well as differences between eyes.

B.
 
8x32 has 4 mm exit pupil;
8x42 has 5.25 mm exit pupil, that is 72% larger by area than 8x32;
8x50 has 6.25 mm exit pupil, that is 42% larger by area than 8x42.

So yes objectively the step in night sky brightness going from 8x32 to 8x42 is bigger than going from 8x42 to 8x50.

Hi Mark

I can't work out your maths. I make it

32 to 42 = 58% increase (in area)

42 to 50 = 70.5% increase

So going from 42mm to 50mm objective is more noticeable in theory than 32mm to 42mm.
 
Hi Mark

I can't work out your maths. I make it

32 to 42 = 58% increase (in area)

42 to 50 = 70.5% increase

So going from 42mm to 50mm objective is more noticeable in theory than 32mm to 42mm.

I did (5.25 / 4)^2 = 1.72 and (6.25 / 5.25)^2 = 1.42. I don't know what you did, sorry.
 
Pies are round :)

Unnecessary to introduce pi.

40+2 squared is 1604+160= 1764.
32 squared is 1024.

1764 divided by 1024 is about 1723 divided by 1000, so about 72%.
 
Pies are round :)

Unnecessary to introduce pi.

40+2 squared is 1604+160= 1764.
32 squared is 1024.

1764 divided by 1024 is about 1723 divided by 1000, so about 72%.

So you're working on the principle of square exit pupils?

Am I missing something here, as the last time I checked my binoculars I saw a round (circular) exit pupil.
 
What has not been discussed, and which has far greater impact than exit pupils and eye pupils is dark adaptation.

This varies with great complexity.
It depends on age, and on whether one has been in bright sunshine or dull conditions during the previous day.
It depends on health and medication. Antibiotics and numerous medicines affect eye sensitivity.

When twilight and night descend then the chemical effects in our eyes take over.
So switching between binoculars as one dark adapts can give incorrect results.
So can exposure on and off to a light.

When dark adapted, merely looking at a bright object such as Jupiter changes the eye condition.
One can be convinced by comparison, even though actually we are interpreting things wrongly.

Places like the National Physics laboratory in the U.K. and in many other countries carry out tests to set standards. These are accurate unless mistakes are made.
I have had accurate transmission tests carried out by Imperial College and a University lab in Finland and the lab technicians presumably know their jobs.

Individuals using non standard methods and not properly trained will probably not give calibrated results.

As to advertised transmissions. This can be peak transmission or averaged or anything.

Regards,
B.
 
Dipped,

As both exit pupils are round, the only variable is 42 and 32.
Or for area the ratio between 42 squared and 32 squared.

Pi just cancels out.

Regards,
B.

P.S.

Pi r(a) squared divided by Pi r(b) squared is the same as

r(a) squared divided by r(b) squared.

Pi divided by Pi is 1.
 
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To pick up on Binastro's point in post #30 . . .

While the standard formula to calculate the area of a circle produces an exact result, for most to use it requires a calculator of some sort

And while that’s not a big deal with the proliferation of hand held digital devices, there is a much simpler way to calculate the relative area of two or more circles
- just square the diameter

That is, eliminate Pi as it’s a constant, and don’t bother to divide the diameter before squaring it
It's a good technique for when you have to use your own embedded device

Leica uses this simplified calculation in it’s manuals for the relative area of exit pupils, where it’s called Geometric Light Value
And it works equally well to compare the relative area of objectives


John
 

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To pick up on Binastro's point in post #30 . . .

While the standard formula to calculate the area of a circle produces an exact result, for most to use it requires a calculator of some sort

And while that’s not a big deal with the proliferation of hand held digital devices, there is a much simpler way to calculate the relative area of two or more circles
- just square the diameter

That is, eliminate Pi as it’s a constant, and don’t bother to divide the diameter before squaring it
It's a good technique for when you have to use your own embedded device

Leica uses this simplified calculation in it’s manuals for the relative area of exit pupils, where it’s called Geometric Light Value
And it works equally well to compare the relative area of objectives


John

Thanks for this, John. Very helpful!
 
Hi Mark

Apologies - I worked out my %age increase incorrectly.

I agree with you though I would express the increases as 172% and 142% respectively.

Mark is correct because we are talking about the extent of the increase.

A 100% increase would mean double the starting point, because you start from the existing 100% so an increase of 100% takes you to 200%.
A 172% increase would mean double plus another 72% on top or 272% of where you started.

In your case the increase is 'only' 72%.

Lee
 
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I always find a 5mm ep a more comfortable viewing experience than a 4mm whatever the light conditions, and that increased comfort makes me choose the 42mm more often than not. I`v learned to never undervalue a comfortable view.
 
Moose8,

Try using the 8X42 for a while (say a week) without the 8X32, then go back to the 8X32.

I am also in agreement with the aforementioned opinion of a 5mm ep, unless it is a good porro prism glass.

Andy W.
 
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