T
thomasstyron
Hello:
I am trying to incorporate an IF Logic function into my worksheet bu
it doesn’t seem to work.
I am working on a body fat calculator, and one of the components o
this calculator is the gender of the individual. I have a drop dow
menu in a cell, with Male and Female for choices. This is in cell B2.
In cell B9, I have the formulas for Body Density and Body fa
incorporated together.
They are as follows:
1. Body Density (BD): 1.109380 - [0.0008267 x (SKF)] + [0.0000016
(SKF)2] - (0.0002574 x Age)
2. Body Fat (in percent): [(4.95/Db) - 4.5] x 100
(From Jackson & Pollock (1978))
Written for Excel, I have condensed it to:
((4.95/(1.10938-(0.0008267*B8)+(0.0000016*((B8)^2))-(0.0002574*A2)))-4.5)
I omitted the *100 at the end, as the cell that the result is displaye
in is formatted for percent.
This formula works great, with cell B8 being the sum of thre
measurements (in mm) on the body and cell A2 being the age. The result
are displayed as a percent.
Unfortunately, this formula is applicable for males only. For females
the constants are changed, but the formula remains essentially th
same.
The female formula is:
((5.01/(1.0994921-(0.0009929*B8)+(0.0000023*((B8)^2))-(0.0001392*A2)))-4.57)
(Derived from the formulas of Jackson et al., (1980))
This formula too works well.
Now I am trying to incorporate the two functions into one functio
using an IF Logic; this is where I hit a snag.
I have come to here but without success:
=IF(B2=Male,"((5.01/(1.0994921-(0.0009929*B8)+(0.0000023*((B8)^2))-(0.0001392*A2)))-4.57))","((4.95/(1.10938-(0.0008267*B8)+(0.0000016*((B8)^2))-(0.0002574*A2)))-4.5)")
This is in cell B9; again cell B8 is the sum of three body measurement
and cell A2 is the age.
My expectation was that if Male is selected from the pull down menu i
cell B2, then the first formula is used. If not then the latter formul
for Females is used. Unfortunately I have been disappointed and hav
become frustrated. The functio
(=IF(B2=Male,"((5.01/(1.0994921-(0.0009929*B8)+(0.0000023*((B8)^2))-(0.0001392*A2)))-4.57))","((4.95/(1.10938-(0.0008267*B8)+(0.0000016*((B8)^2))-(0.0002574*A2)))-4.5)")i
displayed in the cell B9 rather than the computed value.
I have used “ “, [ ], for the separation of values; and I have trie
substituting 1 and 2 for Male and Female (no indicator of superiority
to simplify matters.
If anyone can offer help in this matter I would greatly appreciate it
Thank you for your time.
Best Regards,
Thomas Styro
I am trying to incorporate an IF Logic function into my worksheet bu
it doesn’t seem to work.
I am working on a body fat calculator, and one of the components o
this calculator is the gender of the individual. I have a drop dow
menu in a cell, with Male and Female for choices. This is in cell B2.
In cell B9, I have the formulas for Body Density and Body fa
incorporated together.
They are as follows:
1. Body Density (BD): 1.109380 - [0.0008267 x (SKF)] + [0.0000016
(SKF)2] - (0.0002574 x Age)
2. Body Fat (in percent): [(4.95/Db) - 4.5] x 100
(From Jackson & Pollock (1978))
Written for Excel, I have condensed it to:
((4.95/(1.10938-(0.0008267*B8)+(0.0000016*((B8)^2))-(0.0002574*A2)))-4.5)
I omitted the *100 at the end, as the cell that the result is displaye
in is formatted for percent.
This formula works great, with cell B8 being the sum of thre
measurements (in mm) on the body and cell A2 being the age. The result
are displayed as a percent.
Unfortunately, this formula is applicable for males only. For females
the constants are changed, but the formula remains essentially th
same.
The female formula is:
((5.01/(1.0994921-(0.0009929*B8)+(0.0000023*((B8)^2))-(0.0001392*A2)))-4.57)
(Derived from the formulas of Jackson et al., (1980))
This formula too works well.
Now I am trying to incorporate the two functions into one functio
using an IF Logic; this is where I hit a snag.
I have come to here but without success:
=IF(B2=Male,"((5.01/(1.0994921-(0.0009929*B8)+(0.0000023*((B8)^2))-(0.0001392*A2)))-4.57))","((4.95/(1.10938-(0.0008267*B8)+(0.0000016*((B8)^2))-(0.0002574*A2)))-4.5)")
This is in cell B9; again cell B8 is the sum of three body measurement
and cell A2 is the age.
My expectation was that if Male is selected from the pull down menu i
cell B2, then the first formula is used. If not then the latter formul
for Females is used. Unfortunately I have been disappointed and hav
become frustrated. The functio
(=IF(B2=Male,"((5.01/(1.0994921-(0.0009929*B8)+(0.0000023*((B8)^2))-(0.0001392*A2)))-4.57))","((4.95/(1.10938-(0.0008267*B8)+(0.0000016*((B8)^2))-(0.0002574*A2)))-4.5)")i
displayed in the cell B9 rather than the computed value.
I have used “ “, [ ], for the separation of values; and I have trie
substituting 1 and 2 for Male and Female (no indicator of superiority
to simplify matters.
If anyone can offer help in this matter I would greatly appreciate it
Thank you for your time.
Best Regards,
Thomas Styro