How to determine tilt

[OPTIDONN]

Is there a way to accurately determine the degree of pantoscopic or retroscopic tilt in a pair of glasses?

[chip anderson]

If you want to be exact get a protractor equiped level. Available at any hardware or home supply.

[HarryChiling]

You could take a picture from the side, (camera being level on a tripod or something then measure the degree of tilt) newer didgital camaras are very cheap making it a viable option for someone to turn into a measuring device. Maybe even equip it as part of those machines that show you frames and stuff. As for face form tilt you could do the same, take a picture from the top and measure it. I would hope at some point someone can come along and invent a tool for this.

[chip anderson]

Now that I think about it, some where in my dispensing table I think I have a freebie plastic devise that has such divisions on it. If I can find it I will pass on the name on it.

Chip

Of course you get into the discussion: Should the angle be as opposed to 90 degrees from gravitation force or x degrees from the patient's head if it were held level, or 90 degrees from the angle at which the patient actually holds his head when errect, or even when sitting reading. Now that I have intensly clouded the issue I shall retire for the evening.

[Darryl Meister]

Keep in mind that you are talking about measuring frame tilt here. Lens tilt (for power compensation) will also depend upon the base curve and any decentration or prism, since the optical axis will be tilted relative to the plane of the frame.

[David Wilson]

Both Rodenstock and Zeiss produce little tools to measure tilt. I suspect that this may be what you have Chip. Zeiss's gauge is a little square plastic plate with a needle weighted on one end. The edge of the plate is placed against the lens when the person is in thei habitual posture and the weight6, acting as a plumb-bob.The needle then points to the amount of tilt. It will not take into consideration the points raised by Darryl. Nor wsill Rodenstock's gauge, which is a rather neat little device. It clips onto the temple and the gauge is a u shaped curve with a ball-bearing. The ball-bearing rests at the angle of the tilt.

Another way of measuring tilt has been recommended by Rodenstock and others. It involves finding the progressive fitting height (pupil centre in the habitual posture). Then find the aspheric fitting height, by setting pantoscopic tilt at zero. This is done by getting the wearer to tilt their head until the lens is vertical. Determine pupil centre height at this position. The pantoscopic tilt then is twice the separation of these two markings. That is, if the distance between the aspheric height and the progressive height is 5mm then the pantoscopic tilt is 10 degrees. This is based on the rule that the optical centre for aspherics must be dropped 1mm for every 2 degrees of pantoscopic tilt.
Regards
David

[Darryl Meister]

This is based on the rule that the optical centre for aspherics must be dropped 1mm for every 2 degrees of pantoscopic tilt

Yeah, it's basically a mathematical simplification. At a typical stop distance (to the center of rotation) of 27 mm, an angle of 1 degree translates to a tangent of 0.47 mm across the lens plane.

[lensgrinder]

Lens tilt (for power compensation) will also depend upon the base curve and any decentration or prism, since the optical axis will be tilted relative to the plane of the frame.

Is there a formula that takes this in account? If so, can you please explain?

At a typical stop distance (to the center of rotation) of 27 mm, an angle of 1 degree translates to a tangent of 0.47 mm across the lens plane.

Is stop distance the same as vertex distance?


Thanks in advance

[David Wilson]

Is stop distance the same as vertex distance?

No, the 27mm includes an assumed back vetex distance of 12mm plus the distance from the pole of the cornea to the centre of rotation of the eye of 15mm.

Regards
David

[HarryChiling]

Basically what is being said is that you are geting a reading of the height at the as worn position (with tilt) and the position when the lens is perpidicular with the optical axis. The equation is

tangent (tilt) = difference in height / center of rotation to the back of the lens

in the example in my picture

tan (x) = 5/25
tan (x) = 0.2
x = 11.31 degrees

Simple way of determining tilt I like it.

[lensgrinder]

Thanks Harry and David.
What part does base curve, decentration, and/or prism play when it comes to tilt? For example, if a frame is determined to have 15 degrees of face form with 5mm of decentration o.u., will the face form change with each base curve and if so how do you determine this?

Thanks again.

[William Walker]

Not to sound dense, but how do any of these methods (including the last one) indicate anything other than frame tilt? If base curve or decentration matters, I don't see where there's any room for these to be variables.

Thanks,
Just soaking it up

[Darryl Meister]

Decentering a meniscus lens actually tilts the optical axis of the lens, since the optical center is actually passing along an arc. This should be considered for horizontal (wrap) tilt. You should also consider Martin's rule and the position of the optical centers relative to the pupil when calculating vertical (panto) tilt.

[HarryChiling]

Darryl correct me if I am wrong but this is where prism would be helpfull in the compensation.

sin (angle) = decentration / thickness
decentration = sin (angle) * thickness

then put the decentration into prentices rule to get your prism.

so if a +5.00 lens is tilted 20 degrees 2.0mm center thickness

decentration=sin(20)*2
decentration=0.68
then in prentices rule

prism=5*0.068
prism=0.34

I know that the equation should be more complicated than this and am working on it. I would venture to say that if you were to put a ray of light through the lens snell's law would aaply and change the anlge the light exits the lens therefore changing the angle used in my compensation. I have used this calculation and done well with it so far, it is by no means a polished calculation.

[lensgrinder]

Greetings Harry,

I do not mean to jump in, but I found a formula that I think answers your question. It takes the base curve and thikness of the lens into account.

p=100*(d/n)*(F1)*(angle)

where:

p is prism
d is thikness in meters
n is refractive index
F1 is Base curve
angle is face form in radians

So using 20 degrees of tilt(0.35 rad), 4.5mm thickness, 1.5 index, and an 8.00 base lens. This lens would produce 0.84D of prism BO. So you would grind 0.84D of BI prism.

I hope this helps.


I am still curious though, what part base curve plays in overall tilt as stated in the post above so if you or any one else has insight to this it would be very helpful.

Thanks

[Darryl Meister]

I do not mean to jump in, but I found a formula that I think answers your question. It takes the base curve and thikness of the lens into account.

Do you have a copy of my Introduction to Ophthalmic Optics or something? That particular equation isn't very common...

Harry is referring to the prism introduced by the slight horizontal displacement that results from tilting a lens (assuming you keep the pole of at least surface directly in front of the pupil). The equation you've cited determines the approximate amount of prism introduced by the tilt, itself (ignoring the effects of decentration). You would actually use both, but keep in mind that they are approximations.

[Darryl Meister]

As for the tilt introduced by decentration, again, it occurs because of decentering a meniscus lens. For argument's sake, assume that you are using a guided mini-bevel on the lens that follows the curve of the front surface. If the lens has a flat front surface, decentering the optical center of the lens will simply translate the optical axis horizontally. However, if the front surface is curved, decentering the optical center will actually move the pole over the arc of the front surface. Since the optical axis is perpendicular to this pole, it is effectively tilted. (If anyone has a copy of Stimson's Ophthalmic Dispensing, there is a figure demonstrating this effect on page 178.) I worked out a derivation of this for my program, but it would take me a bit to extract it back out of the code. Maybe I'll have a look at it this weekend.

[lensgrinder]

Do you have a copy of my Introduction to Ophthalmic Optics or something? That particular equation isn't very common...

It came from Optometry and Vision Science written by Ralf Blendowske.

[Darryl Meister]

It came from Optometry and Vision Science written by Ralf Blendowske.

Oh, good. I got the formula from a lens designer about 7 years ago, but I have never confirmed the derivation. I guess I can assume now that he didn't just make it up. ;)

[HarryChiling]

Any way one of you can send me that article?

[HarryChiling]

Wow Darryl the article gives a slightly more precise equation for tilts than Keatings and does mention and work keatings equations in the article. Thanks lensgrinder

[Darryl Meister]

Yeah, I think he uses the geometric mean to calculate mean power instead of the arithmetic mean. Unfortunately, the differences are too small to allow me to allow me to verify them one way or the other with ray-tracing software.

[HarryChiling]

I sent you the article, it does mention that ray trcing will not pick up on the slight differences. All the same it is an interesting approach. The article does mention that for you to obtain a 0.25D difference you would have to be working with 20D in power, all the patients I have seen that have that high of a correction just want vision and high wrap frames don't even come to mind.