object displacement in bifocals

[slackware]

hi all, im an apprentice optician and there's a few questions i want to ask about the above topic. im having a little trouble finding the obj. displacement from the distance o.c. to the near o.c. for example id do it in a ft 40 (40x25) the readling level would be set at 12 and it would be 4 below the distance o.c. to the seg. i know how to plot the prism formula, but at times it's difficult to understand how the professor stated how to find the near prism. is there an easier way to solve for this? thanks.

[HarryChiling]

Think of it as two seperate lenses and then add the prisms together to get the total prism power.

For example your FT40 lets say an Add of +2.00 and the Rx along the 90 is -4.00

doc to seg = 4mm
doc to noc = 12mm
Prism = Power x dec_cm

So our first lens would be:

Prism = -4.00 x 1.2cm = -4.8

our second lens would be 12mm - 4mm = 8mm

Prism = +2.00 x 0.8cm = 1.6

Now we add those two together and whalla

Prism at Near OC = -4.8 + 1.6 = -3.2D

[lensgrinder]

It sounds like you are looking for image jump since you mentioned displacement, but I am not sure.

You need to find out where the OC of the segment is.

You said that the width is 40mm and the depth 25mm.



So your seg_oc is 5mm

To find image jump you use Prentice's Rule using the 5 mm times the add power.
For example a +2.00 Add with this seg would produce 1^BD of image jump.

Now you also mentioned the reading level and how far above the seg your OC is.
Now the top of your seg is 4mm below the OC, your reading level is 12 mm and you now know that the seg_oc is 5mm below the top of the seg.
(5 + 4) - 12 = -3
The negative indicates that you are going lower than the seg_oc.

If you have a -3.00 X 90 with a +2.00 Add
Calculate prism at distance and near
P_distance = 3.00 X 1.2
P_distance = 3.6^BD
P_near = 0.3 X 2 (I use 3mm because the difference between the seg_oc and how far below this the readers eye is)
P_near = 0.6^BU (BU because it is below the seg_oc)

So you have a total of 3.0^BD at reading level.
I attached an image to show the relation between the seg_oc, distance OC and where the reader is reading.

[HarryChiling]

It sounds like you are looking for image jump since you mentioned displacement, but I am not sure.

You need to find out where the OC of the segment is.

You said that the width is 40mm and the depth 25mm.



So your seg_oc is 5mm

To find image jump you use Prentice's Rule using the 5 mm times the add power.
For example a +2.00 Add with this seg would produce 1^BD of image jump.

Now you also mentioned the reading level and how far above the seg your OC is.
Now the top of your seg is 4mm below the OC, your reading level is 12 mm and you now know that the seg_oc is 5mm below the top of the seg.
(5 + 4) - 12 = -3
The negative indicates that you are going lower than the seg_oc.

If you have a -3.00 X 90 with a +2.00 Add
Calculate prism at distance and near
P_distance = 3.00 X 1.2
P_distance = 3.6^BD
P_near = 0.3 X 2 (I use 3mm because the difference between the seg_oc and how far below this the readers eye is)
P_near = 0.6^BU (BU because it is below the seg_oc)

So you have a total of 3.0^BD at reading level.
I attached an image to show the relation between the seg_oc, distance OC and where the reader is reading.

Missed the Depth of the seg and just assumed the top of the seg, once again Brent you saved my a$$.

[slackware]

thanks for the information. my instructor for my class said when calculating the distance prism, just plug in the true power@90 degrees and x it by the reading level divided by 10. he said it's the easiest thing to do but the near is the one to fight for. but yeah, i was going half of the 40 = 20 -25 =5 from the ft40x25. and the rl is 12 and 4 below the d.o.c. so 12-4=8 and 8-5=3 the image jump is another one too. wouldnt it be the add power x 5 divided by 10? or am i just confusing myself? thanks.

[lensgrinder]

Yes it is the Add power times the Seg OC in centimeters or you can keep in millimeters and divide by 10.

Thanks Harry, but you have saved me a couple of time too.;)

[slackware]

cool. i think i have it down to a science now. i'll put up one problem i have and let the novice examine it. i had to figure out the V.I. and slaboff. should i keep it in this thread or make a new one?