stabiliser accuracy

[bibra]

i have a dataset which can be converted into a graph
how do i get an overlapping graph showing the average of the highs and lows

1 17.02
2 16.42
3 16.54
4 16.11
5 17.08
6 17.3
7 17.24
8 -8.86
9 -10.64
10 -8.34
11 -4.56
12 -0.86
13 0.02
14 2.16
15 3.78
16 4.28
17 1.3
18 1.3
19 0.9
20 4.3
21 4.7
22 10.5
23 5.86
24 3.96
25 5.4
26 6.44
27 7.32
28 8.78
29 9.48
30 10.98
31 20.32
32 21.86
33 22.42
34 23.34
35 24.74
36 24.96
37 21.9
38 19.44
39 19
40 19.24
41 16.86
42 14.62
43 15.64
44 13.64
45 11.3
46 0.44
47 8.52
48 3.64
49 2.32
50 1.18
51 3.24
52 6.2
53 2.54
54 4.22
55 2.94
56 5.96
57 7.28
58 9.75
59 9.22
60 9.78
61 41.7
62 15.4
63 16.78
64 18.62
65 18.6
66 15.72
67 16.28
68 18.44
69 13.4
70 10.56
71 12.58
72 12.68
73 15.32
74 18.9
75 19.48
76 21.66
77 25.25
78 24.06
79 17.8
80 18.22
81 18.76
82 16.32
83 16.44
84 20.48
85 22.1
86 19.64
87 23.14
88 25.42
89 21.58
90 22.94
91 20.96
92 17.12
93 13.74
94 10.5
95 11.44
96 10.08
97 11.1
98 8.84

 

[thisoldman]

Hi Bibra,

I still don't understand your "average of the highs and lows." It is not at all clear to me.

average is where the peaks go from max--h1 to min--h2
h1+h2/2=b1

I asked for an example using the short column of cells in post No.4. I need a better explanation.

Your replied the maximum was 17.24. I will agree.
You stated the minimum was -10.64. I will agree with that also.

Then you stated that the first point in column B would be:
=17.24+10.64/2 =14.14

Here is why I'm confused:
That equation doesn't calculate to 14.14!
You've changed the sign of the minimum value from negative to positive!
I want to believe the Excel formula you want is:

=(ABS(MAX(B1:B7)) + ABS(SMALL(B1:B7, 1))) / 2


You stated the next value in column B can be calculated by the equation:
=10.64+.86/2
You didn't calculate the results.
Would this next Excel formula give the correct result?

=(ABS(SMALL(B!:B7, 1)) + ABS(SMALL(B1:B7, 2))) /2


You stated the third value should be the result of this equation:
.86+4.28/2
Where in the heck did that 4.28 come from?

 

[bibra]

Hi Bibra,

I still don't understand your "average of the highs and lows." It is not at all clear to me.


I asked for an example using the short column of cells in post No.4. I need a better explanation.

Your replied the maximum was 17.24. I will agree.
You stated the minimum was -10.64. I will agree with that also.

Then you stated that the first point in column B would be:
=17.24+10.64/2 =14.14

Here is why I'm confused:
That equation doesn't calculate to 14.14!
You've changed the sign of the minimum value from negative to positive!
I want to believe the Excel formula you want is:

=(ABS(MAX(B1:B7)) + ABS(SMALL(B1:B7, 1))) / 2


You stated the next value in column B can be calculated by the equation:
=10.64+.86/2
You didn't calculate the results.
Would this next Excel formula give the correct result?

=(ABS(SMALL(B!:B7, 1)) + ABS(SMALL(B1:B7, 2))) /2


You stated the third value should be the result of this equation:
.86+4.28/2
Where in the heck did that 4.28 come from?

why don't we start again

let me just say what is prescribed in the procedure

draw the original graph(consists of around 1000 points)
draw a graph connecting the mean (/2) points
draw a second mean graph
now calculate the mils as follows

from each peak of the original graph drop a vertical intercept onto the second mean line and note the displacement(A)
then the mils is given as under

=(A1+A2+A3.......An)/(n*65) =should not be greater than 0.8 mils

n=number of peaks
65=constant based on the focal length of camera, magnification etc

procedure sheet with the drawing is uploaded

https://www.dropbox.com/s/hj7xsvitfs...edure.bmp?dl=0

https://www.dropbox.com/s/00idfse06n...lines.png?dl=0

 

[thisoldman]

Bibra, the Dropbox links don't work. I get a 404 message.

 

[bibra]

Bibra, the Dropbox links don't work. I get a 404 message.

https://www.dropbox.com/s/hj7xsvitfsc0g34/procedure.bmp?dl=0

https://www.dropbox.com/s/00idfse06nuyx3z/hand drawn chart with 2 mean lines.png?dl=0

 

[thisoldman]

Bibra,

All I get are the two pictures of the drawn charts. There is no procedure given for how the two mean lines are calculated or drawn. What's needed is the method to determine the means.

 

[bibra]

Bibra,

All I get are the two pictures of the drawn charts. There is no procedure given for how the two mean lines are calculated or drawn. What's needed is the method to determine the means.

we need to add the peak values and divide by 2

(h1+h2)/2

 

[thisoldman]

Bibra, I know how to take an average. It's the selection of h1 and h2 that is still not clear. If I follow your last directions, post #16 , I get one point. I've asked several times, but you have not given a clear answer.