Finishing Room Calculations Flashcards
Given a frame with the following specs:
A= 56
B=48
ED = 60
DBL = 19
Patient PD distant 36/35 near 34/33
Seg height ordered = 23
Find Decentration, inset, total decentration, vertical decentration/seg drop, and minimum blank size
The first thing to do is to find frame PD, which is A+DBL = 56+19 = 75. 75/2= 37.5
Decentration: OD: 37.5 - 36 = 1.5mm IN; OS: 37.5 - 35 = 2.5mm IN
Inset: Can be found by subtracting the near PD from the distant PD, so OD: 36 - 34 = 2mm IN; OS: 35 - 33 = 2mm IN
Total decentration: Inset + Decentration, so OD = 2 + 1.5 = 3.5mm IN; OS = 2 + 2.5 = 4.5mm IN
Vertical decentration: B/2-seg, so OU = 48/2 - 23 = 24 - 23 = 1mm
Minimum Blank Size: ED + 2(decentration) + 2, so OD: 60 + 2(1.5) + 2 = 65mm; OS: 60+2(2.5)+2 = 67mm
Given a frame with the following specs:
A= 52
B=44
ED = 60
DBL = 16
Patient PD distant 33/32 near 30/29
Seg height ordered = 26
Find Decentration, inset, total decentration, vertical decentration/seg drop, and minimum blank size
The first thing to do is to find frame PD, which is A+DBL = 52+16 = 68. 68/2= 34
Decentration: OD: 34 - 33 = 1mm IN; OS: 34 - 32 = 2mm IN
Inset: Can be found by subtracting the near PD from the distant PD, so OD: 33 - 30 = 3mm IN; OS: 32 - 29 = 3mm IN
Total decentration: Inset + Decentration, so OD = 3 + 1 = 4mm IN; OS = 3 + 2 = 5mm IN
Vertical decentration: B/2-seg, so OU = 44/2 - 26 = 22 - 26 = 4mm UP
Minimum Blank Size: ED + 2(decentration) + 2, so OD: 60 + 2(1) + 2 = 64mm; OS: 60+2(2)+2 = 66mm
Given a frame with the following specs:
A= 56
B=50
ED =62
DBL = 18
Patient PD distant 31/33 near 30/32
Seg height ordered = 20
Find Decentration, inset, total decentration, vertical decentration/seg drop, and minimum blank size
The first thing to do is to find frame PD, which is A+DBL = 56+18 = 74. 74/2= 37
Decentration: OD: 37 - 31 = 6mm IN; OS: 37 - 33 = 4mm IN
Inset: Can be found by subtracting the near PD from the distant PD, so OD: 31 - 30 = 1mm IN; OS: 33 - 32 = 1mm IN
Total decentration: Inset + Decentration, so OD = 6 + 1 = 7mm IN; OS = 4 + 1 = 5mm IN
Vertical decentration: B/2-seg, so OU = 50/2 - 20 = 25 - 20 = 5mm
Minimum Blank Size: ED + 2(decentration) + 2, so OD: 62 + 2(6) + 2 = 76mm; OS: 62+2(4)+2 = 72mm
Given a frame with the following specs:
A= 50
B=44
ED = 56
DBL = 20
Patient PD distant 33/31 near 30/29
Seg height ordered = 20
Find Decentration, inset, total decentration, vertical decentration/seg drop, and minimum blank size
The first thing to do is to find frame PD, which is A+DBL = 50+20 = 70. 70/2= 35
Decentration: OD: 35 - 33 = 2mm IN; OS: 35 - 31 = 4mm IN
Inset: Can be found by subtracting the near PD from the distant PD, so OD: 33 - 30 = 3mm IN; OS: 31 - 29 = 2mm IN
Total decentration: Inset + Decentration, so OD = 2 + 3 = 5mm IN; OS = 4 + 2 = 6mm IN
Vertical decentration: B/2-seg, so OU = 44/2 - 20 = 22 - 20 = 2mm
Minimum Blank Size: ED + 2(decentration) + 2, so OD: 56 + 2(2) + 2 = 62mm; OS: 56+2(24)+2 = 66mm
Given a frame with the following specs:
A= 48
B=44
ED = 52
DBL = 19
Patient PD distant 36/35 near 34/33
Seg height ordered = 23
Find Decentration, inset, total decentration, vertical decentration/seg drop, and minimum blank size
The first thing to do is to find frame PD, which is A+DBL = 48+19 = 67. 67/2= 33.5
Decentration: OD: 33.5 - 36 = 2.5mm OUT; OS: 33.5 - 35 = 1.5mm OUT
Inset: Can be found by subtracting the near PD from the distant PD, so OD: 36 - 34 = 2mm IN; OS: 35 - 33 = 2mm IN
Total decentration: Inset + Decentration, so OD = 2.5 OUT + 2 IN = 0.5mm OUT; OS = 1.5 OUT + 2 IN = 0.5mm IN
Vertical decentration: B/2-seg, so OU = 44/2 - 23 = 22 - 23 = 1mm UP
Minimum Blank Size: ED + 2(decentration) + 2, so OD: 52 + 2(2.5) + 2 = 59mm; OS: 52 + 2(1.5)+2 = 57mm
