Formula to turn a Fraction to Decimal

"Easy"? I don't know if it is "easy", but, you just need to extract the different parts from the text string and convert them to appropriate numbers.

=VALUE(LEFT(text,1)) will return the 1 foot. The trick with extracting this will be knowing how many digits to return. Will the feet always be 1 digit? Can it be 2 digits? Do you need the formula to figure out how many digits for the feet part of the text string, or can you always trust that it will be however many digits (using leading zeros when needed)?

=VALUE(MID(text,6,7)) will extract the 1.6875 inches. Again, the real difficulty is will this always be 7 characters long, or should it be 8 characters long to account for the 10 and 11 inch increments (and force you to enter as 01 11/16) or do you need the formula to figure out on the fly how many characters to extract?

Of course, once you have the inches, you can convert to feet by dividing by twelve, then adding the initial feet number. =VALUE(LEFT(text,1)+VALUE(MID(text,6,7))/12 will convert that text string to 1.140625. From there, it's just a matter of knowing if that will always be the right way to parse the text string, or if you need the formula to figure out what to use for the 1, 6, and 7 parts of those text formulas. If this were my thing, and I assume that I will always be under 99 feet, I make the data entry step so that it will always be 2 digit feet and 2 digit inches and 2 digits for the fraction parts (aa' - bb cc/dd") with leading zeros when needed (your example would be entered as 01' - 01 11/16"), and use something like =VALUE(LEFT(text,2))+VALUE(MID(text(7,8))/12

Are you understanding what the 1 and 2 are doing in the LEFT() function? The 6,7 and 7,8 in the MID() function? It should be the same function, just adjusting these parameters for the new text template. If you are not familiar with the LEFT() and MID() functions: https://support.office.com/en-us/art...6-1ea52b99640c and https://support.office.com/en-us/art...8-4ecb12433028
The new text template would be aaaa' - bb cc/dd", and the new function would be =VALUE(LEFT(text,4))+VALUE(MID(text,9,8))/12

In a different direction, I would be inclined to do a text to columns to separate the feet from the inches, and, if I could do it at the same time, get rid of the apostrophe and quotation mark characters.

So text to columns using - as the delimiter would get the feet and inches into separate columns. Something like VALUE(SUBSTITUTE(feet_text,"'","")) to get the feet number standing alone as a number. Then something like =VALUE(SUBSTITUTE(inches_text,CHAR(34),"")) to get the inches value alone as a number. Then a simple summation =feet_value+inches_value/12. Done that way, you end up with 5 additional columns (2 for the text to columns output, and 3 helper columns to convert the text to numbers and add them together). That might be the "easiest" way to do it.

So much depends on what kind of solution you need.

Give this one a try
Enter formula in B1 and copy down

Formula:

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v	A	B
1	0' - 2 3/8"	2.875
2
3	0' - 6 1/16"	3.8125
4
5	1' - 0 1/2"	1.5
6
7	1' - 6 1/4"	16.25
8
9	1' - 8 1/4"	21.25
10
11	1' - 11 15/16"	70.6875
12	1' - 11 15/16"	70.6875
13
14	8' 5"
15
16	2' - 1 11/16"	8.9375
17	2' - 1 11/16"	8.9375
18
19	2' - 1 15/16"	9.1875
20	2' - 1 15/16"	9.1875
21	2' - 1 15/16"	9.1875
22
23	2' - 7 3/4"	20.25
24
25	2' - 11 3/8"	16.125
26	2' - 11 3/8"	16.125
27
28	2' - 11 15/16"	71.6875
29	2' - 11 15/16"	71.6875
30
31	3' - 1 1/8"	4.375
32
33	3' - 11 3/4"	31.25
34
35	4' - 0"
36
37	4' - 0 9/16"	4.5625
38
39	6' - 5 1/2"	31.5

See if this one will give you better results

Formula:

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v	A	B
1	0' - 2 3/8"	2.375
2	1' - 1 11/16"	13.6875
3	0' - 6 1/16"	6.0625
4
5	1' - 0 1/2"	12.5
6
7	1' - 6 1/4"	18.25
8
9	1' - 8 1/4"	20.25
10
11	1' - 11 15/16"	23.9375
12	1' - 11 15/16"	23.9375
13
14	8' 5"
15
16	2' - 1 11/16"	25.6875
17	2' - 1 11/16"	25.6875
18
19	2' - 1 15/16"	25.9375
20	2' - 1 15/16"	25.9375
21	2' - 1 15/16"	25.9375
22
23	2' - 7 3/4"	31.75
24
25	2' - 11 3/8"	35.375
26	2' - 11 3/8"	35.375
27
28	2' - 11 15/16"	35.9375
29	2' - 11 15/16"	35.9375
30
31	3' - 1 1/8"	37.125
32
33	3' - 11 3/4"	47.75
34
35	4' - 0"	48
36
37	4' - 0 9/16"	48.5625
38
39	6' - 5 1/2"	77.5

There are some of them that can't be resolved like: 1' - 7" or 8' 5" At least I couldn't find a solution for them yet.

The solution I proposed in post #6 (text to columns followed by a couple of substitute() formulas) works for those cases -- it just doesn't all fit into a single cell.

Would that work, or must it be a single cell mega formula?

Then I guess (at least for me) that the answer to the OP question is "No, there is not an easy function."

I haven't got time to piece a complete function together, but I could see adding the FIND() function to these text functions to help separate feet from inches https://support.office.com/en-us/art...3-d0d89b0a0628

LEFT(text,FIND("'",text)-1) should return the feet number.
Something like MID(text,FIND("-",text)+2,FIND(CHAR(34),text)-FIND("-",text)-2) should find the inches number.
Then, those two variations of LEFT() and MID() fit into the summation like above =VALUE(LEFT(...))+VALUE(MID(...))/12
I haven't got time right now to test and fine tune those FIND() functions to make sure the correct stuff will always be returned, but one should be able to coax something like that to give you the result you want.