Lets try it a different way, for 11.3 what answer do you expect?

"HARSHAWARDHAN. S .SHASTRI" wrote:

Thanks Mike and Bernard for quick response.

iIhave tried both formula's on my data but i am getting diff up to 110
microns.

Any other logic / formula. ?

H S Shastri

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"Mike H" wrote:

Hi,

It much depends on what happens between 11 & 12 degrees this assumes the
change is linear

=((LOOKUP(C2,A2:B69)-LOOKUP(CEILING(C2,1),A2:B69))/10*MOD(C2,1)*10)+LOOKUP(C2,A2:B69)

with 11.3 in C2 and your table in columns A&B

Mike

"HARSHAWARDHAN. S .SHASTRI" wrote:

I AM WORKING ON CAM DATA.WHAT I WANT IS TO FIND INTERMIDIATE VALUE OF A
CURVE.
FOLLOWING IS MY DATA ,WHAT I NEED IS TO FINF THE VALUE SAY AT 11.3 DEG.

DEG LIFT
0.0 9.0000
1.0 9.0000
2.0 9.0000
3.0 9.0000
4.0 9.0000
5.0 9.0000
6.0 9.0000
7.0 9.0000
8.0 8.9874
9.0 8.9539
10.0 8.8972
11.0 8.8154
12.0 8.7089
13.0 8.5770
14.0 8.4185
15.0 8.2318
16.0 8.0151
17.0 7.7655
18.0 7.4795
19.0 7.1500
20.0 6.7883
21.0 6.4133
22.0 6.0454
23.0 5.6961
24.0 5.3647
25.0 5.0498
26.0 4.7507
27.0 4.4664
28.0 4.196
29.0 3.9388
30.0 3.6941
31.0 3.4613
32.0 3.2396
33.0 3.0286
34.0 2.8279
35.0 2.6368
36.0 2.4549
37.0 2.2819
38.0 2.1173
39.0 1.9609
40.0 1.8122
41.0 1.671
42.0 1.5366
43.0 1.4099
44.0 1.2895
45.0 1.1755
46.0 1.0678
47.0 0.9661
48.0 0.8704
49.0 0.7802
50.0 0.6957
51.0 0.6165
52.0 0.5426
53.0 0.4738
54.0 0.4099
55.0 0.3511
56.0 0.2971
57.0 0.2477
58.0 0.2032
59.0 0.163
60.0 0.1274
61.0 0.0963
62.0 0.0697
63.0 0.0473
64.0 0.0294
65.0 0.0156
66.0 0.0063
67.0 0.0011