| Home |
| Search |
| Today's Posts |
|
#1
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
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 |
|
#2
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
Try using the FORECAST function
With you data in A1:B69, I used =FORECAST(11.3,B12:B14,A12:A14) giving me a result of 8.78 best wishes -- Bernard V Liengme Microsoft Excel MVP http://people.stfx.ca/bliengme remove caps from email "HARSHAWARDHAN. S .SHASTRI" wrote in message ... 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 |
|
#3
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
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 |
|
#4
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
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 ================================================== ======= "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 |
|
#5
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
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 ================================================== ======= "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 |
|
#6
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
Hi,
Now I've correct my (long winded) formula I get the same answer as Bernard, but next time I'll use Forecast :) 8.78345 =(LOOKUP(C2,A2:B69))-(((LOOKUP(C2,A2:B69))-LOOKUP(CEILING(C2,1),A2:B69))/10)*(MOD(C2,1)*10) Mike "Mike H" wrote: 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 ================================================== ======= "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 |
|
#7
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
This is now bugging me. The interval between (11) 8.8154 & (12) 8.7089 =
-0.01065 and can be shown thus:- 11.00 8.81540 11.10 8.80475 11.20 8.79410 11.30 8.78345 11.40 8.77280 11.50 8.76215 11.60 8.75150 11.70 8.74085 11.80 8.73020 11.90 8.71955 12.00 8.70890 So why doesn't the correct answer for 11.3 = 8.78345 or does the curve demonstrate some strange profile between these 2 values that we haven't been made aware of. Mike "Mike H" wrote: 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 ================================================== ======= "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 |
|
#8
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
I think part of the problem is that you are assuming a linear relationship
between points when the curve for the posted data looks sort of like the right half of a "bell curve" (shape-wise, sort of an elongated backward "S"). Since the OP wants 1 micron "precision" (6th decimal place for data to 5 decimal places?), I'm guessing the curving part of the curve is not close enough to a straight line to give the precision the OP is looking for. -- Rick (MVP - Excel) "Mike H" wrote in message ... This is now bugging me. The interval between (11) 8.8154 & (12) 8.7089 = -0.01065 and can be shown thus:- 11.00 8.81540 11.10 8.80475 11.20 8.79410 11.30 8.78345 11.40 8.77280 11.50 8.76215 11.60 8.75150 11.70 8.74085 11.80 8.73020 11.90 8.71955 12.00 8.70890 So why doesn't the correct answer for 11.3 = 8.78345 or does the curve demonstrate some strange profile between these 2 values that we haven't been made aware of. Mike "Mike H" wrote: 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 ================================================== ======= "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 |
|
#9
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
Rick,
I think your probably correct but I did give the OP the opportunity to clarify that in my first post It much depends on what happens between 11 & 12 degrees this assumes the change is linear and got no such clarification/correction Mike "Rick Rothstein" wrote: I think part of the problem is that you are assuming a linear relationship between points when the curve for the posted data looks sort of like the right half of a "bell curve" (shape-wise, sort of an elongated backward "S"). Since the OP wants 1 micron "precision" (6th decimal place for data to 5 decimal places?), I'm guessing the curving part of the curve is not close enough to a straight line to give the precision the OP is looking for. -- Rick (MVP - Excel) "Mike H" wrote in message ... This is now bugging me. The interval between (11) 8.8154 & (12) 8.7089 = -0.01065 and can be shown thus:- 11.00 8.81540 11.10 8.80475 11.20 8.79410 11.30 8.78345 11.40 8.77280 11.50 8.76215 11.60 8.75150 11.70 8.74085 11.80 8.73020 11.90 8.71955 12.00 8.70890 So why doesn't the correct answer for 11.3 = 8.78345 or does the curve demonstrate some strange profile between these 2 values that we haven't been made aware of. Mike "Mike H" wrote: 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 ================================================== ======= "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 |
|
#10
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
The data fits a Gompertz curve with this formula fairly well:
=ka*EXP(-EXP(kb-kc*xdata))+kd where ka -9.21E+00 kb 2.17E+00 kc 9.25E-02 kd 9.11E+00 The sum of the squared absolute error = .764 which is not good enough for the precision required. If we consider only the 7 data points around 11 and apply a second order polynomial, the results might be good enough: =TREND(ydata,xdata^{1,2},11.3^{1,2}) =8.78598 |
|
#11
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
Hi Harshawardhan,
I did a trial with 2 polynomial trends, one from 7 Deg to 21 Deg, the other from 21 Deg to 67 Deg. I had to go right up to a 6th order polynomial to improve the accuracy of the curve fit. I don't think it is quite as tight as you are looking for but it is very close, plug some different values into the 'new X' cell and you will see what I mean. Actually, I thought that I should have gotten a better fit so maybe someone else can take a look and see if I have got a small error in there somewhere. You can download a copy here http://www.savefile.com/files/1833930 It's in XL2000 format. HTH Martin "HARSHAWARDHAN. S .SHASTRI" wrote in message ... 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 |
|
#12
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
Hi again,
I just used Autocad to calculate the true value. At 11.3 Deg your Y value will be 8.785991 My spreadsheet calcs this value to be 8.786205 A difference of 0.000214 Can you live with that variation? I am confident that the Autocad derived result is correct, however it is not a straightforward procedure that can be used as a generic solution, it is just a one off type of thing. HTH Martin "MartinW" wrote in message ... Hi Harshawardhan, I did a trial with 2 polynomial trends, one from 7 Deg to 21 Deg, the other from 21 Deg to 67 Deg. I had to go right up to a 6th order polynomial to improve the accuracy of the curve fit. I don't think it is quite as tight as you are looking for but it is very close, plug some different values into the 'new X' cell and you will see what I mean. Actually, I thought that I should have gotten a better fit so maybe someone else can take a look and see if I have got a small error in there somewhere. You can download a copy here http://www.savefile.com/files/1833930 It's in XL2000 format. HTH Martin "HARSHAWARDHAN. S .SHASTRI" wrote in message ... 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 |
|
#13
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
Thank u Mike ,
You have taken many efforts and really that works.The formula which you have modified working very well and gives me precision up to 6 microns which is acceptable to be. Thanks Once again H S Shastri ================================================== ========= "Mike H" wrote: This is now bugging me. The interval between (11) 8.8154 & (12) 8.7089 = -0.01065 and can be shown thus:- |
|
#14
Posted to microsoft.public.excel.worksheet.functions
|
|||
|
|||
|
INTERMIDIATE VALUE OF A CURVE (PRECISION 1 MICRON)
Hi,
I seem to be talking to myself down here but I will continue as I think it is important. As your data is to 4 decimal places and you are talking microns I am assuming that your lift column is in centimetres making 0.0001 equal to 1 micron. I have proved to myself, with the use of Autocad, that the true Y value at X=11.3 is 8.7860 (rounded to 4 decimals) My Spreadsheet returns 8.7862 (out by 2 microns) The straight line value of 8.7835 that you say is only out by 6 microns is, in fact, out by 25 microns. Of course, I may be stumbling with some false assumptions here, as you haven't provided a great deal of detail, but it appears to me that your calculations may be flawed. HTH Martin "MartinW" wrote in message .. . Hi again, I just used Autocad to calculate the true value. At 11.3 Deg your Y value will be 8.785991 My spreadsheet calcs this value to be 8.786205 A difference of 0.000214 Can you live with that variation? I am confident that the Autocad derived result is correct, however it is not a straightforward procedure that can be used as a generic solution, it is just a one off type of thing. HTH Martin "MartinW" wrote in message ... Hi Harshawardhan, I did a trial with 2 polynomial trends, one from 7 Deg to 21 Deg, the other from 21 Deg to 67 Deg. I had to go right up to a 6th order polynomial to improve the accuracy of the curve fit. I don't think it is quite as tight as you are looking for but it is very close, plug some different values into the 'new X' cell and you will see what I mean. Actually, I thought that I should have gotten a better fit so maybe someone else can take a look and see if I have got a small error in there somewhere. You can download a copy here http://www.savefile.com/files/1833930 It's in XL2000 format. HTH Martin "HARSHAWARDHAN. S .SHASTRI" wrote in message ... 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 |
| Reply |
| Thread Tools | Search this Thread |
| Display Modes | |
|
|
Similar Threads
|
||||
| Thread | Forum | |||
| fractions with more precision | Excel Discussion (Misc queries) | |||
| Precision in formulas? | Excel Worksheet Functions | |||
| Higher precision: can one get it? | Excel Discussion (Misc queries) | |||
| curve fitting a charging capacitor type curve | Excel Discussion (Misc queries) | |||
| Precision displayed does not match precision in cell | Excel Discussion (Misc queries) | |||
![HARSHAWARDHAN. S .SHASTRI[_2_] is offline](https://www.excelbanter.com/images/SultanThemeVB3R/statusicon/user_offline.gif)
Linear Mode
