Are plus lenses heavier than minus?

[Robert Martellaro]

There is a thread over at sci.med.vision (05-18-05 Re: Farsighted v. Nearsighted Ratio) that turned into a discussion of the advantages and disadvantages of being hyperopic or myopic, with one person declaring that with all things being equal (edge and CT, power, design, material etc), plus is heavier than minus. Another person thought minus would be heavier than plus. I ran a series of powers through my dispensing software and plus was about 20% heavier in low to medium powers, with the differential increasing with higher powers. I'm *guessing* that a meniscus lens is more efficient, mass wise, with minus powers, compared to plus. Two questions...

1. Is my software describing the situation correctly?

2. If it is, why is plus heavier?

Thanks

[chip anderson]

Plus has to have added center thickness in order to have enough thickness at the farthest extremity from the eye (optical) center. This will increase greatly with large frames and some shapes such as tear-drop.

Minus lenses increase in edge thickness for the opposite reason but do not need anything added at the center to make them strong enough for the edge mount at the thinnest point.

Chip

[Robert Martellaro]

Chip,

My example was using cr39 with a 2.0 CT for the minus and 2.0mm ET for the plus, round 60mm with no DEC, using best form BCs. In other words a level playing field. Still, the minus powers keep coming up lighter than plus powers. Any further thoughts for this optical theory challenged optician?

Thanks

[35oldguy]

Using the math method if you have a +2.00 D lens with a ET of 2mm you will have a CT of 3.8 CT with zero decentration. Or with a -2.00 D lens 2.0 CT you will have a 3.8 ET. Weight difference? Have you put them on a scale to see if there is a difference? I wonder?!!!

Chip,

My example was using cr39 with a 2.0 CT for the minus and 2.0mm ET for the plus, round 60mm with no DEC, using best form BCs. In other words a level playing field. Still, the minus powers keep coming up lighter than plus powers. Any further thoughts for this optical theory challenged optician?

Thanks

[DrG]

Since I am the person who made this claim on sci.med.vision, here is my reasoning:

Represent a simple plus lens as a hemisphere with radius of R.

Represent a simple minus lens as a cylinder containing the same
hemisphere as above, where the height of the cylinder is the same as R,
and the area of the base is represented by pi times the radius squared.
The volume of the cylinder is pi*R^2*h or pi*R^3. Subtracting the
volume of the sphere from the volume of the cylinder gives the volume
of the simple minus lens with a CT of zero. Compare this to the volume
of the simple plus lens with ET of zero.

If I use 2 as the value of R in the example, I come up with 16.75 for
the volume of the hemisphere = plus lens, and I come up with
25.12-16.75 = 8.37 for the volume of the minus lens. For those who are math challenged, use 1 for the value of R.

Please check my math.

DrG

[chip anderson]

Frames, optical centers and P.D.s make the playing field very convoluted.

[Robert Martellaro]

Frames, optical centers and P.D.s make the playing field very convoluted.

True. But this discussion concerns two identical lens configurations except for power.

Cr39 using best form base curves, 2mm CT for minus, 2mm ET for the plus, no bevel, identical shape. My program gives me these numbers-

60mm diameter -1.00DS weighs 8.53g and 10.57g in plus
60mm diameter -3.00DS weighs 10.13g and 15.3g in plus
60mm diameter -7.00DS weighs 13.82g and 26.1g in plus

DrG,

Your math is fine. I like the idea of determining the volume of the different shapes. Here's a link to a nice math tool...

http://grapevine.abe.msstate.edu/~fto/tools/vol/ ).

Now, is there a tool that will calculate the weight taking into consideration the complex curves of Rx meniscus lenses?

Thanks for your help,

[Darryl Meister]

65 mm Corrected Curve Hard Resin Blank, 1 mm Edge

+4.00 D = 15.1 g
+2.00 D = 9.5 g
+1.00 D = 6.9 g

65 mm Corrected Curve Hard Resin Blank, 2 mm Center

-1.00 D = 11.4 g
-2.00 D = 13.9 g
-4.00 D = 18.8 g

65 mm Corrected Curve Hard Resin Blank, 1 mm Center

-1.00 D = 7.0 g
-2.00 D = 9.4 g
-4.00 D = 14.4 g

Best regards,
Darryl

[Darryl Meister]

Now, is there a tool that will calculate the weight taking into consideration the complex curves of Rx meniscus lenses?

http://www.optiboard.com/forums/showthread.php?t=12650

Best regards,
Darryl

[DrG]

65 mm Corrected Curve Hard Resin Blank, 1 mm Edge

+4.00 D = 15.1 g
+2.00 D = 9.5 g
+1.00 D = 6.9 g

65 mm Corrected Curve Hard Resin Blank, 2 mm Center

-1.00 D = 11.4 g
-2.00 D = 13.9 g
-4.00 D = 18.8 g

65 mm Corrected Curve Hard Resin Blank, 1 mm Center

-1.00 D = 7.0 g
-2.00 D = 9.4 g
-4.00 D = 14.4 g

Best regards,
Darryl

It's that easy now, is it? OK, edge them all down to a 60mm eye, geometrically centered, and reweigh.

DrG

[Darryl Meister]

OK, edge them all down to a 60mm eye, geometrically centered, and reweigh.

60 mm Corrected Curve Hard Resin Blank, 1 mm Edge

+4.00 D = 11.3 g
+2.00 D = 7.4 g
+1.00 D = 5.5 g

60 mm Corrected Curve Hard Resin Blank, 1 mm Center

-1.00 D = 5.6 g
-2.00 D = 7.4 g
-4.00 D = 10.9 g

And these are calculated values from a program I wrote, not measured values. Generally, the volume of a plus lens will be very similar to the volume of a minus lens of the same power when both have the same minimum thickness and lens form.

Best regards,
Darryl

[DrG]

It would seem that we have a conundrum between programs. Since my theoretical calculations predict that the plus lens of the same radius as the minus lens has more mass than the minus lens, I would have to side with Robert.

The only resolution would seem to be handguns at 6 paces, or actual laboratory scale measurements.

DrG

[Darryl Meister]

It would seem that we have a conundrum between programs. Since my theoretical calculations predict that the plus lens of the same radius as the minus lens has more mass than the minus lens, I would have to side with Robert.

Below is the volume calculation of a +4.00 D plano-convex lens and a -4.00 D plano-concave lens at a 60 mm diameter with zero minimum thickness in hard resin. Feel free to double-check the math.

Assumptions

D = Diameter of lens blank (60 mm)
F = Power of major surface (4.00 D)
R = Radius of curvature of major surface
N = Refractive index (1.500)
S = Sagitta of major surface

R = 1000 * (n - 1) / F
R = 1000 * (1.500 - 1) / 4.00 = 125.0 mm

S = R - SQRT(R^2 - (D/2)^2)
S = 125 - SQRT(125^2 - 30^2) = 3.65 mm

Volume of 4.00 Section of a Sphere

Vs = PI/3 * S^2 (3R - S)
Vs = PI/3 * 3.65^2 (3 * 125 - 3.65) = 5180.81 mm^3 = 5.18 cm^3

This represents the volume of a +4.00 D plano-convex spectacle lens at a diameter of 60 mm with a knife edge thickness.

Volume of Cylinder

Vc = PI * (D/2)^2 * S
Vc = PI * 30^2 * 3.65 = 10320.13 mm^3 = 10.32 cm^3

Net Volume of Cylinder - Section of a Sphere

Vn = Vc - Vs
Vn = 10.32 - 5.18 = 5.14 cm^3

This represents the volume of a -4.00 D plano-concave spectacle lens at a diameter of 60 mm with a zero center thickness.

Note that there is virtually no difference between these two lens volumes.

Best regards,
Darryl

[OPTIDONN]

Below is the volume calculation of a +4.00 D plano-convex lens and a -4.00 D plano-concave lens at a 60 mm diameter with zero minimum thickness in hard resin. Feel free to double-check the math.

Assumptions

D = Diameter of lens blank (60 mm)
F = Power of major surface (4.00 D)
R = Radius of curvature of major surface
N = Refractive index (1.500)
S = Sagitta of major surface

R = 1000 * (n - 1) / F
R = 1000 * (1.500 - 1) / 4.00 = 125.0 mm

S = R - SQRT(R^2 - (D/2)^2)
S = 125 - SQRT(125^2 - 30^2) = 3.65 mm

Volume of 4.00 Section of a Sphere

Vs = PI/3 * S^2 (3R - S)
Vs = PI/3 * 3.65^2 (3 * 125 - 3.65) = 5142.90 mm^3 = 5.14 cm^3

This represents the volume of a +4.00 D plano-convex spectacle lens at a diameter of 60 mm with a knife edge thickness.

Volume of Cylinder

Vc = PI * (D/2)^2 * S
Vc = PI * 30^2 * 3.65 = 10320.13 mm^3 = 10.32 cm^3

Net Volume of Cylinder - Section of a Sphere

Vn = Vc - Vs
Vn = 10.32 - 5.14 = 5.18 cm^3

This represents the volume of a -4.00 D plano-concave spectacle lens at a diameter of 60 mm with a zero center thickness.

Note that there is virtually no difference between these two lens volumes.

Best regards,
Darryl

Here we just trust Darryl. The man works for Sola and boy can he make us feel stupid real quick (in a good way:) ). The guy is a walking optical program!!

[Darryl Meister]

The guy is a walking optical program!!

Years of moderating the Ophthalmic Optics Forum will do that to you... ;)

Best regards,
Darryl

[HarryChiling]

The math is like poetry, this is not all opthalmic optics but is applied that way Bravo.

[Darryl Meister]

Okay, I was feeling a bit inspired, so I decided to prove mathematically exactly why a plano-convex lens is roughly equal in volume to a plano-concave lens of the same diameter and power.

Recall that the volume of a section of a sphere is given by,

Vs = PI/3 * S^2 (3R - S)

which represents a plano-convex lens with a zero edge thickness.

Since S (the sag) will usually be small in relation to the radius of curvature R, this formula can be simplified using,

Vs = PI/3 * S^2 * 3R

Now recall that to calculate the volume of a plano-concave lens, we must first determine the volume of a cylinder of equivalent thickness using,

Vc = PI * (D/2)^2 * S

We then subtract the volume of the section of a sphere from the volume of this cylinder in order to determine the volume of a plano-concave lens with a zero center thickness,

Vn = Vc - Vs

Moreover, if we want to determine the difference in volume between a plano-concave lens and a plano-convex lens, you would subtract the volume of the plano-convex lens from the volume of the plano-concave lens,

Dif = Vn - Vs = (Vc - Vs) - Vs = Vc - 2 * Vs

Now, we substitute for Vc and Vs,

Dif = Vc - 2 * Vs
Dif = PI * (D/2)^2 * S - 2 * PI/3 * S^2 * 3R

Now, again since the sag S is usually small relative to the radius or curvature R, we can substitute this approximate sag formula for S,

S = (D/2)^2 / (2R)

Finally, we substitute and then simplify,

Dif = PI * (D/2)^2 * S - 2 * PI/3 * S^2 * 3R
Dif = PI * (D/2)^2 * (D/2)^2 / (2R) - 2 * PI/3 * [(D/2)^2 / (2R)]^2 * 3R
Dif = PI * (D/2)^4 / (2R) - 2 * PI * R * (D/2)^4 / (4R^2)
Dif = PI * (D/2)^4 / (2R) - PI * (D/2)^4 / (2R)
Dif = 0

This proves mathematically that when the radius of curvature R is relatively long compared to the sagitta S of the surface, a plano-convex lens produces no difference in volume from a plano-concave lens of the same power and diameter.

Best regards,
Darryl

[DrG]

According to your "approximate" formula, your answer is "approximately" true. :hammer:

DrG

[OPTIDONN]

DrG do you have a version of this equation? If you do please share. I like to copy them down and try to learn them. The more the merrier!:cheers:

[DrG]

I don't have my own version, but I can certainly oblige. How many versions would you like?
:drop:

DrG

[OPTIDONN]

I'll take anything and everything!

[Darryl Meister]

According to your "approximate" formula, your answer is "approximately" true.

But ultimately true enough to prove that there is no real advantage to plus versus minus power when it comes to lens volume -- at least until you start factoring in things like minimum thickness differences. For our +/-4.00 lens at 60 mm, even the exact difference is less than 0.8%.

Best regards,
Darryl

[DrG]

But ultimately true enough to prove that there is no real advantage to plus versus minus power when it comes to lens volume -- at least until you start factoring in things like minimum thickness differences. For our +/-4.00 lens at 60 mm, even the exact difference is less than 0.8%.

Best regards,
Darryl

Theoretically, there is a difference, even if it is less than one percent for your example. In practice, there are thickness differences, as the thicker parts of the minus lens are discarded during edging. But, as you have shown, there are ways of minimizing those differences.

DrG

[Darryl Meister]

In practice, there are thickness differences, as the thicker parts of the minus lens are discarded during edging.

Very true, especially as the frame difference increases. But also remember that minus lenses generally have more minimum thickness than plus lenses (e.g., a 2.0 CT versus a 1.0 ET), which will offset some of this. Still for a shaped lens you could probably argue that the plus lens would be heavier, though the actual magnitude of the difference would depend upon several factors.

Best regards,
Darryl

[Darryl Meister]

For a typical frame (52 x 35 mm) with 3 mm of decentration, the volumes of the shaped lenses would be.

-4.00 D: 4.39 cm^3


+4.00 D: 5.07 cm^3


Best regards,
Darryl