Please help on basic ABO study questions. (Finding Rx from given powers)

[moody]

Hello, I am just starting to study for ABO exam and have been stuck on trying to figure out how to find Rx when powers are given at certain axis. I can find powers from given Rx, but just going backwards, I just can't grasp the concept so if anyone can explain to me I'd really appreciate it.
For example, what is the Rx when -2.00@37 and -3.00@127?
I can go as far as figuring out that the cylinder power would be 1.00, (difference between two powers) but not even sure whether - or +, because from 37 to 127, it'd be -1.00 but from 127 to 37, it'd be +1.00. Which do I use for the Rx??
And then what is the next step?? take the half of that 1.00 and add it to either -2.00 or -3.00 to get the sph equivalent at either axis or is that not necessary? do I just write the -2.00 then the found cylinder power(- 1.00) at 37? and same for -3.00 +1.00 @ 127? Then which Rx do I write for the answer? Do I need to write both?


For example 2, what is the Rx when -1.00@130 and Plano@85?
This is more confusing because it's not 90 degrees apart, I can't even figure out the cylinder power straight. So do I use 30-45-60 rule to figure out the cylinder power at 40 first? (since 40 is 90 away from 130)? Now then, I take +1.00 multiply that by 0.5 since 85 is 45 degrees away from 130, and get +0.5 for new cylinder power at 40, right?
Then the same question as above, do I just write out the new cylinder power and given powers at given degrees as a correct Rx? so would it be -1.00+0.5@130 or am I completely wrong? I'm just so confused.. Please help.

[Uncle Fester]

Your @ for axis and the rx is confusing.

At least to me but that's not hard to do!

Where are they coming from?

Try posting again with an rx's we can read.

Remember first and foremost the full power of cylinder is 90 degrees from the axis.

[wmcdonald]

The Rxs based on the cross cyls are clearly written. The OP has correctly provided the Rxs in bith + and - cyl.

Either is correct, but since most are - cyl that is recommended.

Hope this is helpful

[moody]

Your @ for axis and the rx is confusing.

At least to me but that's not hard to do!

Where are they coming from?

Try posting again with an rx's we can read.

Remember first and foremost the full power of cylinder is 90 degrees from the axis.

Thank you for taking the time to read my post. I found these questions online trying to practice for the exam. Anyways,
the problems were worded as they were written. Original questions said

"2) What is the Rx of a lens if the power is -2.00 @037 and -3.00 @127
4) What is the RX of a lens if the power is -1.00 @130 & PL @085?"

[moody]

The Rxs based on the cross cyls are clearly written. The OP has correctly provided the Rxs in bith + and - cyl.

Either is correct, but since most are - cyl that is recommended.

Hope this is helpful

Thank you, so you are saying there is no need for spherical equivalent for both of the problems, I just need to write them out as is? That's good. But I still need help because for the second problem the answer says it is "+1.00 -2.00 040 or -1.00 +2.00 130"
And I don't get 2.00 I got 0.5 as the cylinder power. What am I missing?

[drk]

Draw an "X" (you will need a protractor with what's called "TABO" notation:
https://www.2020mag.com/article/power-analysis
https://www.2020mag.com/article/the-...ion-and-vision


When they say "at" (as opposed to saying "axis", which is what we really do, which is harder, but that's optics for you) technically it means just to put the power on the line that it says.

For example -9.00 @ 90 means put that power up at the top. -4.00 @ 180 means put that power on the side.

After constructing that "optical cross", THEN and only then do you draw out whatever prescription form that is asked for. Cool people like me use minus cylinder notation. Stuffy losers like ophthalmologists use plus cylinder notation. Weirdos from France use their own snobby notation. Only the English-speaking world does it right, and of that group, American Optometrists are the King of Beers (and American Opticians are Bud Light).

[wmcdonald]

I would be pleased to have a conversation to explain this but doing it here will be time consuming. To attempt to be brief, you were provided 2 cross cylinders. In the 037 Meridian the power is -2.00, correct? In the 127 the power is -3.00. To determine the Rx, pick one. If you choose -2.00, then you must add -1.00 to reach the total of -3.00 in that meridian. So, the Rx is -2.00-1.00 x 027. If you use simple flat transposition, the Rx in plus is -3.00 +1.00 x127. Same Rx either way.

Th questions doesn't ask for spherical equivalent the way you present it, it asks for Rx.

Again, I hope this is helpful.

[Uncle Fester]

The Rxs based on the cross cyls are clearly written. The OP has correctly provided the Rxs in bith + and - cyl.

Either is correct, but since most are - cyl that is recommended.

Hope this is helpful

Never saw an @ trying to resolve cylinder power @ another meridian when I was in school but that was before the earth was accepted as round.

(Okay- an oblate spheroid;)

[moody]

@wmcdonald,

Thank you for clarifying it for me. I guess using the "optical cross" as drk mentioned, seeing it on the cross and with your explanation, I see what's going on.

Thanks for taking the time to reply to this post!

[wmcdonald]

The protractor Dr K mentions is not an Optical Cross. I assume you are an Apprentice?

[Opticraft97]

Laramy K on youtube has some good videos for visual representation if youre a visual learner. might help! Maybe start with "Laramy K optical cross" on YouTube!

[KrystleClear]

Laramy K on youtube has some good videos for visual representation if youre a visual learner. might help! Maybe start with "Laramy K optical cross" on YouTube!

The Optical Cross videos on there helped me a lot!