# Help needed to determine operating focal length

Started by
mike bacanin
, Apr 08 2013 07:11 AM

11 replies to this topic

### #1

Posted 08 April 2013 - 07:11 AM

I could do with your help in determining my operating focal length.

I have a C6,fl 1500mm. I use a 50mm long sct visual back, and into that a Baader 60 degree binoviewer with about 125mm optical path.no diagonal needed with this binoviewer.

any help greatly appreciated

Thanks

Mike

I have a C6,fl 1500mm. I use a 50mm long sct visual back, and into that a Baader 60 degree binoviewer with about 125mm optical path.no diagonal needed with this binoviewer.

any help greatly appreciated

Thanks

Mike

### #2

Posted 08 April 2013 - 12:21 PM

[EDITED TO CORRECT A SLOPPY MATHEMATICAL ERROR]

It's best to measure via the star drift method. Obtain a calliper with which to measure the eyepiece field stop on something like a 20-25mm Plossl, to an accuracy of 0.1mm or better. Find a star within a few degrees of the equator and time how long it takes to traverse the field, in seconds, with the drive turned off.

The field of view in degrees = drift_seconds / 239.3

The focal length = field-stop / TAN(degrees)

It's best to measure via the star drift method. Obtain a calliper with which to measure the eyepiece field stop on something like a 20-25mm Plossl, to an accuracy of 0.1mm or better. Find a star within a few degrees of the equator and time how long it takes to traverse the field, in seconds, with the drive turned off.

The field of view in degrees = drift_seconds / 239.3

The focal length = field-stop / TAN(degrees)

### #4

Posted 08 April 2013 - 05:04 PM

Thank you, very helpful indeed.

Mike

Mike

### #5

Posted 08 April 2013 - 05:53 PM

Due to the sensitivity of focus position on the working focal length, it still might be best to determine the value for *your* setup by actual measurement...

Then you can measure the exit pupil (to 0.05mm with a measuring loupe) in order to calculate the actual f/ratio, in order to verify if full aperture is being utilized. Alternately, you can simply magnify the exit pupil, then place a ruler, or any straight object, across the front end and locate the points which meet at the edge of the exit/entrance pupil, measuring their separation to find the working aperture.

Then you can measure the exit pupil (to 0.05mm with a measuring loupe) in order to calculate the actual f/ratio, in order to verify if full aperture is being utilized. Alternately, you can simply magnify the exit pupil, then place a ruler, or any straight object, across the front end and locate the points which meet at the edge of the exit/entrance pupil, measuring their separation to find the working aperture.

### #6

Posted 08 April 2013 - 06:08 PM

The field of view in degrees = drift_seconds / 31,017,600

Glen, shouldn't that 31,017,600 be a little closer to 240?

Eric

### #7

Posted 08 April 2013 - 10:04 PM

[EDITED TO FIX SLOPPY MATH...]

That big number takes into account the number of seconds in a sidereal day (86,160), times 360 degrees. The number of seconds required for an equatorial star to cross the field diameter, divided by this figure, supplies the field width in degrees.

[In edit]

That large figure is my mistake. Instead of applying 86,160 * 360 = 32 million-ish, it should have been 86,160 / 360 = 239.3.

That big number takes into account the number of seconds in a sidereal day (86,160), times 360 degrees. The number of seconds required for an equatorial star to cross the field diameter, divided by this figure, supplies the field width in degrees.

[In edit]

That large figure is my mistake. Instead of applying 86,160 * 360 = 32 million-ish, it should have been 86,160 / 360 = 239.3.

### #8

Posted 08 April 2013 - 10:48 PM

I must be missing something here. So if the drift time was 120 seconds the field width would be?

Eric

OK I get it,finally. You meant to divide 86,160 by 360 instead of multiplying, thus giving the number of seconds per degree.

Eric

OK I get it,finally. You meant to divide 86,160 by 360 instead of multiplying, thus giving the number of seconds per degree.

### #9

Posted 09 April 2013 - 02:51 AM

Beatlejuice,

You were missing nothing! I worked out the numbers and derived a correct answer when punching it in on my calculator. But I mistake you applied the wrong o

You were missing nothing! I worked out the numbers and derived a correct answer when punching it in on my calculator. But I mistake you applied the wrong o

### #10

Posted 09 April 2013 - 02:55 AM

Beatlejuice,

You were missing nothing! I worked out the numbers and derived a correct answer when punching it in on my calculator. But I mistakenly applied the wrong operand when writing out the formula 'in my head.' I should have trusted my gut when seeing that huge number as I typed it out here.

I'm glad you persisted; your correct operand clued me in to my blunder.

Cheers!

You were missing nothing! I worked out the numbers and derived a correct answer when punching it in on my calculator. But I mistakenly applied the wrong operand when writing out the formula 'in my head.' I should have trusted my gut when seeing that huge number as I typed it out here.

I'm glad you persisted; your correct operand clued me in to my blunder.

Cheers!

### #11

Posted 09 April 2013 - 03:15 AM

You sure had me scratching my head Glen because you so seldom if ever err in these matters.

BTW thanks for "The focal length = field-stop / TAN(degrees)" I didn't know that and will put it to use soon.

Eric

BTW thanks for "The focal length = field-stop / TAN(degrees)" I didn't know that and will put it to use soon.

Eric

### #12

Posted 09 April 2013 - 05:09 AM

Eric,

Using the tangent function is technically wrong, as it relates an angle to, well, a tangent. Angular measure on the sky, and for purposes of magnification, must work on an arc, and so is properly based on the radian. However, at small angles typical of telescopic fields, the error introduced by the tangent function is so small as to be well below the other measurement errors.

I know Jon Isaacs likes his simple (non trig-based) formula built around the figure 57.3, which is the number of degrees in one radian. I derive

TFoV = 57.3 / ( Obj. f.l. / field stop diam.)

No scientific calculator required.

Using the tangent function is technically wrong, as it relates an angle to, well, a tangent. Angular measure on the sky, and for purposes of magnification, must work on an arc, and so is properly based on the radian. However, at small angles typical of telescopic fields, the error introduced by the tangent function is so small as to be well below the other measurement errors.

I know Jon Isaacs likes his simple (non trig-based) formula built around the figure 57.3, which is the number of degrees in one radian. I derive

TFoV = 57.3 / ( Obj. f.l. / field stop diam.)

No scientific calculator required.