Optical center punch formula

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SirJohn

Active Member
Nice to see that you are industrious to make your own. I don't have the same patience and just ordered an optical punch set from Flex Bar for \$53.00.

Scott_M

Well-Known Member
It was more an exercise to check the theory. It is a bit large for anything I may use it for. I will make up a smaller one to use. Probably I/4" dia. punch and element. Also on a skinny base so you can get up close to something. Or a pie shaped base with the punch at the point so you can use it in a corner.

Scott

Stone

Member
Thanks Scott.
Is there any magnification of the image?

Scott_M

Well-Known Member
Is there any magnification of the image?

Yes, I am guessing about 2X ( Based on the 2X in the formula )

Scott

ajoeiam

Well-Known Member
Hmmmmmmmmmm I've been wondering about even the need for this 'optical center punch'.

Can't remember if it was recommended procedure but to get the job done for me I would use the following procedure:

1. scribe my lines (doesn't take real heavy lines!!)
2. use a prick punch (can't remember if that's a 45 or a 60 degree included angle but its quite pointed)
easy to 'feel' the lines and very easy to find the intersection point
its important to have the prick punch as close as possible to vertical to the surface
3. use a center punch (again IIRC that's a 90 degree included angle) to make the mark 'more defined'
I usually use a 3/8" (6 mm) hex tool for this function
4. I managed to get my hands on a 'road grader' replaceable blade tooth that is some kind of hardened tool steel
this thing is about 1" in dia and will take a real wallop - - - - the mark - - - well I'm gotten well over 1/8" deep
great for starting with a large drill bit (say 1/2" (12 or 13 mm)) when I want a large drill bit following - say larger than 1-1/2" (35 mm)

its step 2 that makes the difference - - - - - don't need to see the intersection - - - I can 'feel' it

(on the other hand - - - maybe I 'yust be veered' (need a very heavy Cherman haccent!) (sic))

HTH

Stone

Member
Could everyone with a commercial optical punch please measure the radius and length (from top of curve of radius to flat base) of their optical punches and post the results here. This needs to be done accurately as a little out changes the ratio.

I am interested because the commercial ones do not seem to be following the focal length = half the radius of curvature (F = Radius/2).

The one I broke was: L= 46.6mm, radius 18mm, which is:

F =2.59 x R

Which is a much bigger radius

Stone

Member
My friends commercial one is :
48/20 = 2.4 x R

Harpanet13

Wierd Harold
I use an optical punch as an aid when doing metalwork (see attached picture). It is one of my favorite gadgets! I broke the optical rod when I knocked the device of the table. I use it a lot and thought this is a good time for me to finally make my own improved version that does not require I swop the optical rob with the punch each time.

I have a short length of 10mm Perspex rob that I used to make the optical rob from. It worked well but the focus is completely out. I used a 13 mm radius in my attempt, which is the same as the original one I purchased; however, my viewing rod is shorter and would require a different radius to focus correctly.

I was wondering if anyone knows the formula to calculate the focal length so that I cut the curved end to create a focal point at exactly the other end of the rod, where I have the cross hairs marked? I spent a few hours online but can’t seem to find this equation.

The refractive index of the Perspex should be about 1.495.

I tried another optical rod with radius 11 and it was much better.

View attachment 129721

They are quite time intensive to make so I do not want to have to make many at different radii to try find the correct length, also I am running out of Perspex rod. I live in a remote area so driving to buy more Perspex rod is a 200km round trip.

If anyone has any knowledge in this area, I would appreciate your help.
Issue 1 of Model Engineers' Workshop has an article on building an optical center punch which may help.

Stone

Member
Thanks Harpanet13. Managed to find a copy.
The ratio in that article is :
45/25 = 1.8 x R

Interestingly, the author does not describe how he made the optical lens curvature

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