formula for back curve for HI index

**mika** · 09-08-2006, 08:39 AM

Hi, I need help!
first sorry for my english

Now, I know how to calculate the Back Curve B=P-F/(1-(F*t/n)), it function, but when I want it applicat for higher index, the curves are not the same like from the Rx calculator.
f.i. : when I calculate with other index first must by the formula for desired power (n-1)/(1.498-1)*P ??? is this OK??? We use 1.498 as tool index.

...but when you calculate with bigger Power (p=+8.00, n=1.61.....) I dont have the correct value, i mean Back Curve. it is not the same like from the Rx Calculator

I hope that someone understood me;)

**lensgrinder** · 09-08-2006, 09:04 AM

Do not forget that the lens front curve is based on a 1.53 index. For example: A front curve is listed on the box as 6.25 what is the True front curve in 1.60 if the tooling index is 1.498? (6.25*(1.6-1))/(1.498-1) = 7.53, so when you calculate your curves use 7.53 as the front curve.

**HarryChiling** · 09-08-2006, 09:31 AM

I am not sure what calculator you are using, but most software will be set to use the tool index 1.53. If there is a setting in the software that allows you to change the tool curve change that to 1.498. Maybe this helps other than that you have the correct formulas just use them in the correct order

True curve formula

true curve = (index of material - 1 / index of tool - 1) * marked power

Back vertex power = power - true front curve power / (1 - [true front curve power * thickness / index of material])

This should give you the powers that you need.

**mika** · 09-08-2006, 11:01 AM

Perhaps I wrote wrong
I know the true curve, because I measure it with the clock and then is the value (SAG) conversed to diopter (true front curve). by the way what formula converse SAG to diopter? We use clock with sag gauge diameter=40mm, gauge type=bar, ball dia=3.96875mm

We calculate with old Coburn Software RxPII and the tool index is setting on 1.498, also I dont know where is the difference between your formula and our calculator?

Thanks

**lensgrinder** · 09-08-2006, 11:18 AM

You need to convert these readings to the material you are working with. The lens clock and sag gauge are usaully calibrated for index 1.53. When you measure a curve with a lens clock that is not the true curve you need to take that reading and find the true curve.
Dmarked/Dtrue=(tooling index - 1)/(material index -1). Like I said lens clocks and sag gauges are usually 1.53, so that is what you will use for your tooling index when converting to true power from the reading on the lens clock.
Converting SAG to Diopter:
F = (S X 2000 X(n-1))/S^2 X Y^2
F=Surface Power
S=SAG
n=Refractive Index of Sag gauge
Y=Semi-Diamerer(half of the diameter of the gauge)
y= 20mm in your case

**shanbaum** · 09-08-2006, 11:28 AM

Originally Posted by lensgrinder

You need to convert these readings to the material you are working with. The lens clock and sag gauge are usaully calibrated for index 1.53. When you measure a curve with a lens clock that is not the true curve you need to take that reading and find the true curve.

While lens clocks (which indicate diopters of curvature) usually assume an index of 1.53, sag gauges (which indicate millimeters of sagitta) assume no index at all. What you refer to as the "dioptric index of sag gauge" is really the index in which you want the curve expressed.

Also, you didn't take the ball diameter of mika's gauge into account. The formula you provided works for knife-edge ('bar' or 'bell') gauges.

**lensgrinder** · 09-08-2006, 01:29 PM

Originally Posted by shanbaum

While lens clocks (which indicate diopters of curvature) usually assume an index of 1.53, sag gauges (which indicate millimeters of sagitta) assume no index at all. What you refer to as the "dioptric index of sag gauge" is really the index in which you want the curve expressed.

Also, you didn't take the ball diameter of mika's gauge into account. The formula you provided works for knife-edge ('bar' or 'bell') gauges.

Thank you for pointing these thing out. You can always learn.

**Darryl Meister** · 09-09-2006, 03:47 AM

Mika, Is there a particular question or example that you would like to see worked out?

**lensgrinder** · 09-09-2006, 10:41 AM

After Shanbaum told me that you need to take the ball diameter into account, I would love to know how the formula is derived. I know that you need to input the radius of the ball into the equation, but I have tried and failed to figure out how. Any helpl would be appreciated.

**Darryl Meister** · 09-09-2006, 12:39 PM

I know that you need to input the radius of the ball into the equation, but I have tried and failed to figure out how

Check out this thread on the subject.

**lensgrinder** · 09-10-2006, 07:39 PM

Thank you Darryl. Now I need to figure out how it is derived. I have not looked at it much this week end, but I am going to try on Monday. I was thinking something like:
(r + rb)^2 = (y + rb)^2 - (r-s)^2

where rb is the radius of the ball tip

When I factor this out it does not come up correctly. If you or anyone else has any insight it would be greatly appreciated.

**Darryl Meister** · 09-10-2006, 09:11 PM

It would really depend on how the "sagitta" is measured by the bar gauge. (I've never bothered to look into this, so I couldn't say which is the correct or at least the most common approach.) For example, if the sag gauge measurement is based on the distance from the bottom edge of the middle pin, the derivation would be based on the following geometry:

Or, if the sag gauge measurement is based on the distance from the center of the middle pin, the derivation would be based on this geometry instead:

**HarryChiling** · 09-10-2006, 09:51 PM

I was just about to say that Darryl except without the pictures. :) You had a (y+b)^2 where a y^2 should be.

**HarryChiling** · 09-10-2006, 09:53 PM

And Darryl what do you use for your images? They always look proffesionally done.

**Darryl Meister** · 09-10-2006, 10:04 PM

I use Microsoft's integrated drawing tools for quick and dirty illustrations and Adobe Illustrator for more serious stuff.

**HarryChiling** · 09-10-2006, 10:06 PM

I would think that the second derivation would be unnecessary due to the fact that you should always be tangent to the ball tip gauges center pin making it unnecesary to factor that into the equation. But it does make for some good knowledge.

**mika** · 09-11-2006, 02:58 AM

Thanks all
First I must the informations absorb

**mika** · 09-11-2006, 07:08 AM

Once again, true curve
Base on the box is 6.50, with your formula is the true curve=.498/.53*6.50= 6.10.
When I measure lens from this box with lens clock, I reading on the clock 2.60(mm?) and when this value I give in software i become result - diopter= 6.532. Which one of them is right???

Calculator need:
index = 1.498
sag gauge diameter = 40mm
gauge typ = bar ball diameter = 3.96875mm
sag = 2.6 (?mm?)
---------result----------------
radius = 76.23
diopter = 6.532 - I mean this is the true power, because when I use this value in calculator or in the formula for BC=P-F/(1-(F*t/n)), the result is the same, exception by the HI index lenses

**shanbaum** · 09-11-2006, 07:32 AM

Originally Posted by mika

Once again, true curve
Base on the box is 6.50, with your formula is the true curve=.498/.53*6.50= 6.10.
When I measure lens from this box with lens clock, I reading on the clock 2.60(mm?) and when this value I give in software i become result - diopter= 6.532. Which one of them is right???

Calculator need:
index = 1.498
sag gauge diameter = 40mm
gauge typ = bar ball diameter = 3.96875mm
sag = 2.6 (?mm?)
---------result----------------
radius = 76.23
diopter = 6.532 - I mean this is the true power, because when I use this value in calculator or in the formula for BC=P-F/(1-(F*t/n)), the result is the same, exception by the HI index lenses

The "base on the box" is meaningless in this context, like the eye sizes assigned to frames. The surface power of that "6.50" lens could be anything, and in fact, it's more likely to be some number other than 6.50 than it is to be 6.50.

So, if what you're saying is that when you sag the lens, the measurement converts to a curve of 6.532, that's not a problem. Were you to inquire of the manufacturer of the lens, what true curve he is trying to produce, that would be meaningful. If his specification is in the vicinity of 6.53, you're in the ballpark.

**mika** · 09-11-2006, 08:12 AM

Yes I have the true bases from the manufacturer, but I want to know the principle

thanks

**shanbaum** · 09-11-2006, 08:55 AM

Originally Posted by mika

Yes I have the true bases from the manufacturer, but I want to know the principle

thanks

Then start with the true base, not what's marked on the box. Don't be surprised if the curve you derive from the measured sag is a little different from the specified true curve.

I'm not sure what tolerances (with regard to front curves) lens manufacturers are trying to observe these days.

**HarryChiling** · 09-11-2006, 11:40 AM

Originally Posted by mika

the result is the same, exception by the HI index lenses

what is your resoluts and values for the high index lenses?

**Darryl Meister** · 09-11-2006, 11:48 AM

Base on the box is 6.50, with your formula is the true curve=.498/.53*6.50= 6.10

In the United States (I can't speak for Slovakia), the "true curve" on the box from lens manufacturers is generally based on a tooling index of 1.530. So, if your box was for a hard resin lens (n = 1.498-1.500), your calculation would be correct if you wanted to determine the actual refractive power of the surface. Similarly, an index of 1.586 (poly), instead of 1.498, would result in a refractive power of 0.586/0.53*6.50 = 7.19.

**lensgrinder** · 09-11-2006, 03:58 PM

Originally Posted by Darryl Meister

Or, if the sag gauge measurement is based on the distance from the center of the middle pin, the derivation would be based on this geometry instead:

Well after factoring everthing out to you would use:
(r+b)^2 = (r+b-s)^2 + y^2 for a minus lens
2rs = y^2 + s^2 - 2bs
and
(r+b)^2 = (r-b-s)^2 + y^2 for a plus lens
2rs = y^2 + s^2 + 2bs

Thanks Darryl for the picture and helping me out. I love to know how formulas are derived, it helps me get a better understanding

**HarryChiling** · 09-12-2006, 02:27 PM

Originally Posted by lensgrinder

Well after factoring everthing out to you would use:
(r+b)^2 = (r+b-s)^2 + y^2 for a minus lens
2rs = y^2 + s^2 - 2bs
and
(r+b)^2 = (r-b-s)^2 + y^2 for a plus lens
2rs = y^2 + s^2 + 2bs

I think its the other way around the equations for (+) and (-), but you are correct that once you cross the plano powers or the plate height to compensate for the ball gauge you would have to take that value out of that leg of the trianlge instead of adding it in.