Would this be correct?

[HarryChiling]

I have been working out a few problems and something odd struck me:

Sagittal Power = D * [ 1 + (sin^2(theta) / 2n)]

Tangential Power = D * [ (2n + sin^2(theta)) / (2n*cos^2(theta)) ]

Well we could rearrange the tangetial formula to read:

Tangential Power = Sagittal Power / cos^2(theta)

Wouldn't that be a heck of a lot easier to remember?

[drk]

Did you do that right?

You should run a sample equation both ways to make sure they come out the same.

If so, then you have a simple enough formula to actually visualize the relationship between sagittal power and tangential power.

WHATEVER THE HELL THOSE ARE!

[Jacqui]

I have been working out a few problems and something odd struck me:

Sagittal Power = D * [ 1 + (sin^2(theta) / 2n)]

Tangential Power = D * [ (2n + sin^2(theta)) / (2n*cos^2(theta)) ]

Well we could rearrange the tangetial formula to read:

Tangential Power = Sagittal Power / cos^2(theta)

Wouldn't that be a heck of a lot easier to remember?

HUH ?? You say your what hurts??

[HarryChiling]

I have been working out a few problems and something odd struck me:

Sagittal Power = D * [ 1 + (sin^2(theta) / 2n)]

Tangential Power = D * [ (2n + sin^2(theta)) / (2n*cos^2(theta)) ]

Well we could rearrange the tangetial formula to read:

Tangential Power = Sagittal Power / cos^2(theta)

Wouldn't that be a heck of a lot easier to remember?

For instance,
in the sagittal equation we have:

Sagittal Power = D * [ 1 + (sin^2(theta) / 2n)]

what is we said

cucumber = 1 + sin^2(theta) / 2n

now we can rewrite our equation to read:

Sagittal Power = D * cucumber

In the tangential equation we have the formula:

Tangential Power = D * [ (2n + sin^2(theta)) / (2n*cos^2(theta)) ]

If we were to pull out 1/cos^2(theta) from the equation we would have:

Tangential Power = (1 / cos^2(theta)) * D * [ (2n + sin^2(theta)) / (2n) ]

we can further simplify the last part of that equation:

Tangential Power = (1 / cos^2(theta)) * D * [ 1 + (sin^2(theta) / 2n) ]

Now that last part can be replaced with our previous example cucumber:

Tangential Power = (1 / cos^2(theta)) * D * cucumber

And D * cucumber is the Sagittal Power so we could rewrite it as:

Tangential Power = (1 / cos^2(theta)) * Sagittal Power
Tangential Power = Sagittal Power / cos^2(theta)

To me that's a heck of a lot easier to remember than the previous formula and now I can memorize the sagittal formula and the relationship of the tangential power to the sagittal power. I was just wondering why more books don't present it that way, it seems easier to understand when simplified.

PS - no cucumbers were harmed inthe makign of this example.

[Fezz]

Somebody please get Harry a model to build, a sport, or a blasted therapist!


He is screaming for help people!

;):cheers::D:cheers:;)

[HarryChiling]

Somebody please get Harry a model to build, a sport, or a blasted therapist!


He is screaming for help people!

;):cheers::D:cheers:;)

How about a model to fondle. :D I'll take two please.:cheers:

[Crickett13]

I always come to read Harry's posts in here when I start feeling like I may be more than just a reasonably intelligent guy. It brings me back to earth:hammer:

[HarryChiling]

I always come to read Harry's posts in here when I start feeling like I may be more than just a reasonably intelligent guy. It brings me back to earth:hammer:

Dude you had my head spinning with your knowledge of ophthalmic lens processing so don't you even, I should be callign you a process engineering nerd. :D:cheers:

[Crickett13]

Hey as far as I'm concerned if I see the word Theta I assume we are talking about a college frat house.;)

[drk]

Ok, Harry, nice post on the derivation. That's tough to do on a keyboard.

Since you have really simplified the equation, can you characterize sagittal power and tangential power (of a lens?) and what angle theta is?

Merely guessing: sagittal power is refracting ability of a lens with a normal incidence, and tangential power is with an extreme incident angle, and theta is angle from normal?

[Darryl Meister]

Harry, to answer your question more directly, Yes, the derivation is relatively straightforward when you divide the original equations through by 2n. Ultimately, this represents the relationship between Coddington's original equations...

[Darryl Meister]

Ultimately, this represents the relationship between Coddington's original equations...

Though I should add that this relationship is usually expressed as:

Fs = Ft * cos^2 Theta

[Crickett13]

Dude you had my head spinning with your knowledge of ophthalmic lens processing so don't you even, I should be callign you a process engineering nerd. :D:cheers:

Harry I'm changing my business card but I prefer process engineering geek.:p