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Light Cone Width Confirmation?

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#1 MalVeauX

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Posted 23 October 2021 - 11:56 AM

Hey all,

 

I'm not experienced enough in OSLO yet to build a full model, to answer this for myself and won't be home for a few days so I can't use my stuff anyways.

 

So, just seeing if anyone can do a simple confirmation of the width of the light cone at any particular point, using similar triangles calc:

 

I'm modeling a common Celestron C6R, Synta lens, 150mm F8 with a 1200mm focal length. I'm looking for the point in the light cone where it will pass the light cone, without masking it, through the 2" entrance of a typical 2" focuser when wracked all the way in (max back focus).

 

So I'm using:

 

Width = Distance / Focal Length * Aperture

 

So, solving for 50mm light cone width, I'm coming up with it happening at 400mm from plane of focus. Does that sound approximately correct?

 

Application: shortening this OTA (and eventually re-building as truss) and finding where I can put the focuser (a typical 2" GSO crayford) to have maximum back focus without masking the light cone.

Thanks!

 

Very best,



#2 sixela

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Posted 23 October 2021 - 12:38 PM

You also need to account for off axis light bundles, depending on the fully illuminated field you want.
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#3 MalVeauX

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Posted 23 October 2021 - 12:43 PM

You also need to account for off axis light bundles, depending on the fully illuminated field you want.

Hrm,

 

I'm curious what it would be then to have a 12mm illuminated field vs a 6mm illuminated field.

Don't know how to figure that in!

 

Very best,



#4 jimhoward999

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Posted 23 October 2021 - 01:01 PM

Hrm,

 

I'm curious what it would be then to have a 12mm illuminated field vs a 6mm illuminated field.

Don't know how to figure that in!

 

Very best,

The chief ray angle is pretty small so you will be pretty close by just adding the format to distance/F#.  So just add 12mm or 6mm to your calculation and you will be, not exact (you will underestimate slightly since you aren't exactly telecentric)),  but fairly close.  


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#5 MalVeauX

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Posted 23 October 2021 - 01:03 PM

The chief ray angle is pretty small so you will be pretty close by just adding the format to distance/F#.  So just add 12mm or 6mm to your calculation and you will be, not exact (you will underestimate slightly since you aren't exactly telecentric)),  but fairly close.  

Thanks,

 

So maybe just add 12mm to the 50mm for a 62mm width for the calc to be safe for this purpose?

 

Very best,



#6 dan_h

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Posted 23 October 2021 - 01:07 PM

Hrm,

 

I'm curious what it would be then to have a 12mm illuminated field vs a 6mm illuminated field.

Don't know how to figure that in!

 

Very best,

Take it step by step.

Foe a 12mm illuminated image, the light cone tapers from 150mm to 12mm or a total taper of 138mm. It tapers over the 1200mm focal length distance or 138mm/1200mm. 

 

At any point on the light cone, the diameter would be 12mm plus the amount the light cone expands from the focal plane, or 138/1200  X distance from the focal plane.   This gives 12 + ( 138/1200 X D) = 50mm focuser tube diameter. 

 

12+ (138/1200 X D) = 50mm

(138/1200 XD) = 38mm 

D = 38 X1200/138 = 330.43mm for a 12mm fully illuminated image.

 

D = 44 X 1200/144 = 366.66 for a 6mm fully illuminated image. 

 

Hope this helps, 

 

dan



#7 MalVeauX

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Posted 23 October 2021 - 01:10 PM

Take it step by step.

Foe a 12mm illuminated image, the light cone tapers from 150mm to 12mm or a total taper of 138mm. It tapers over the 1200mm focal length distance or 138mm/1200mm. 

 

At any point on the light cone, the diameter would be 12mm plus the amount the light cone expands from the focal plane, or 138/1200  X distance from the focal plane.   This gives 12 + ( 138/1200 X D) = 50mm focuser tube diameter. 

 

12+ (138/1200 X D) = 50mm

(138/1200 XD) = 38mm 

D = 38 X1200/138 = 330.43mm for a 12mm fully illuminated image.

 

D = 44 X 1200/144 = 366.66 for a 6mm fully illuminated image. 

 

Hope this helps, 

 

dan

Ahhh thanks! That solves that. Appreciate it. smile.gif

 

I just don't want to mask the light cone with the focuser entrance (the 2" stalk, wracked all the way in), and thus mask the aperture effectively lowering potential resolution. So do you think its safe to take the 2" focuser entrance to -366mm and not vignette or mask the light cone then?

 

Very best,


Edited by MalVeauX, 23 October 2021 - 01:17 PM.


#8 Jeff B

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Posted 23 October 2021 - 01:22 PM

I use good old Y=MX+B. Which is basically what Dan did.

 

But you also need to know where along the x axis the various "choke points" to the cone are, such as the inlet to the focuser (and any internal stops it may have) and any stops inside the main tube.  

 

I do this for all of my ATM projects, even Newts.

 

Jeff


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#9 MalVeauX

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Posted 23 October 2021 - 01:25 PM

I use good old Y=MX+B. Which is basically what Dan did.

 

But you also need to know where along the x axis the various "choke points" to the cone are, such as the inlet to the focuser (and any internal stops it may have) and any stops inside the main tube.  

 

I do this for all of my ATM projects, even Newts.

 

Jeff

Thanks,

 

Yea just trying not to choke or mask the aperture via cutting out some of the light cone. It's a super basic synta 150mm F8 lens. The focuser is a standard GSO linear bearing 2" focuser, crayford. Just trying to find where the limit is with pushing the focuser into the light cone without masking it at all, if I were to shorten the OTA to get as much back focus as possible in the system, just to know the limits to not cut off too much.

 

Very best,



#10 Garyth64

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Posted 23 October 2021 - 01:30 PM

"So, solving for 50mm light cone width, I'm coming up with it happening at 400mm from plane of focus. Does that sound approximately correct?"

 

Yes, that is correct.

 

However, a 1/2 degree of the skyd will have an image plane diameter of about 11mm, so at 400mm from the image plane, the diameter of the cone will be a little larger.  (A secondary place at 400 mm from the image plane would need to have a diameter of 57mm.)


Edited by Garyth64, 23 October 2021 - 01:57 PM.

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#11 jimhoward999

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Posted 23 October 2021 - 01:40 PM

Take it step by step.

Foe a 12mm illuminated image, the light cone tapers from 150mm to 12mm or a total taper of 138mm. It tapers over the 1200mm focal length distance or 138mm/1200mm. 

 

At any point on the light cone, the diameter would be 12mm plus the amount the light cone expands from the focal plane, or 138/1200  X distance from the focal plane.   This gives 12 + ( 138/1200 X D) = 50mm focuser tube diameter. 

 

12+ (138/1200 X D) = 50mm

(138/1200 XD) = 38mm 

D = 38 X1200/138 = 330.43mm for a 12mm fully illuminated image.

 

D = 44 X 1200/144 = 366.66 for a 6mm fully illuminated image. 

 

Hope this helps, 

 

dan

That geometry is not quite right because is it assumes a downward sloping chief ray angle, whereas, with stop at the primary, it is really upward sloping.



#12 Jeff B

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Posted 23 October 2021 - 01:43 PM

That geometry is not quite right because is it assumes a downward sloping chief ray angle, whereas, with stop at the primary, it is really upward sloping.

Close enough though to build stuff.



#13 MalVeauX

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Posted 23 October 2021 - 02:02 PM

Thanks all for any commentary, very helpful and interesting! Always contention! :p

 

So, just to simplify, what would be the minimum location in the back focus from the focal plan on the 150mm F8 refractor lens, to pass without masking the opening of a 2" focuser's draw tube at max back focus into the tube? Or is it not that simple and I'm missing a critical piece?

 

Very best,



#14 jimhoward999

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Posted 23 October 2021 - 02:37 PM

Thanks all for any commentary, very helpful and interesting! Always contention! tongue2.gif

 

So, just to simplify, what would be the minimum location in the back focus from the focal plan on the 150mm F8 refractor lens, to pass without masking the opening of a 2" focuser's draw tube at max back focus into the tube? Or is it not that simple and I'm missing a critical piece?

 

Very best,

281mm for 12mm field and no vignetting.


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#15 MalVeauX

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Posted 23 October 2021 - 02:39 PM

281mm for 12mm field and no vignetting.

Thanks! Appreciate that. So anything from there on towards focal plane will not vignette or mask the aperture and is safe then.

 

Very best,



#16 jimhoward999

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Posted 23 October 2021 - 03:34 PM

Thanks! Appreciate that. So anything from there on towards focal plane will not vignette or mask the aperture and is safe then.

 

Very best,

yes


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#17 MalVeauX

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Posted 23 October 2021 - 03:51 PM

yes

waytogo.gifwaytogo.gifwaytogo.gif

 

Thanks!

 

Very best,



#18 Bob4BVM

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Posted 23 October 2021 - 10:24 PM

I use the formulas, but before i start cutting stuff i usually draw out the rays, full scale if possible.  That gives me a clear model of vignetting limits all down the OTA, plus is very handy for sizing and spacing the baffles for best function of those.

 

My full-scale for my 6:F5 was easy on rolled paper, .   For yours use a roll of white butcher paper since yours is a longer FL.

 

CS

Bob

 

dwg obj.jpg

 

dwg rear.jpg


Edited by Bob4BVM, 23 October 2021 - 10:26 PM.

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#19 MalVeauX

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Posted 24 October 2021 - 07:02 AM

I use the formulas, but before i start cutting stuff i usually draw out the rays, full scale if possible.  That gives me a clear model of vignetting limits all down the OTA, plus is very handy for sizing and spacing the baffles for best function of those.

 

My full-scale for my 6:F5 was easy on rolled paper, .   For yours use a roll of white butcher paper since yours is a longer FL.

 

CS

Bob

 

attachicon.gifdwg obj.jpg

 

attachicon.gifdwg rear.jpg

Thanks, that really is a good idea!

 

Very best,



#20 dan_h

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Posted 25 October 2021 - 09:32 AM

That geometry is not quite right because is it assumes a downward sloping chief ray angle, whereas, with stop at the primary, it is really upward sloping.

I believe the geometry is correct.

 

For 100% illumination, the shape and size of the light cone is determined by the marginal rays coming from the extreme edge of the aperture. If the marginal rays get through there is no need to consider any other rays unless you have a central obstruction. 

 

dan


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