So.. where's my math error?

[Mike Kolb]

I expect to catch heat for being a mathematical idiot, but here goes:  

Refretted a customer's electric with s/s frets. Ordered the frets (4ft) and it all worked out fine. The end.

Before doing the work, and while standing-around waiting for something to dry on another job, my mind wandered to the frets. The guitar (an Ibanez RG) has 24 frets... so I measured the 1st fret (1.730") and then the 24th fret (2.270").  

My thinking was: by adding the 2 lengths together and dividing by 2 I'd get the average fret length ... which I did and it was exactly 2-inches. 

Then panic set-in. "If the average length is 2" and there's 24 frets, that's 48" and that's the length of the fretwire I ordered, with no room for error whatsoever".

As it turned-out, there was plenty of fretwire left-over... maybe 10" or so. The job's done, the customer's happy.... but where'd I go wrong here?  My better half thinks the supplier may have been generous, but I never measured the actual wire when it was received.  

There's bound to be someone here who'll get a chuckle before setting me straight?

[Mike Kolb]

I didn't intend to demean him:)  The "kid" is actually a pretty good player and it's just that everybody, nowadays, seems like a kid to me!

No.. the guitar didn't need a refret at all... in fact, it was brand-new. But the customer likes stainless frets so much that it's become the first order of business when he gets a new guitar.

[RETROROD]

Funny....as new as that guitar looked?? SSfrets entered my mind> It does seem to be a trend. More 'bang' for the buck, perhaps?

The most ridiculous of that trend is..... I have a "friend" buyer of guitars....primarily to flip "these days".He has a 'refret guru' That he takes to 'said' "guru" to have re-fretted with stainless steel. The MOST aggregious (sp) offense being a sweet..... "late 60's" Martin with great original frets.

Jeeesh!

[Robbie Collins]

Be glad he isn't using the neon red DR Strings.

[Andrew]

I have a set of those on my NJ Stealth! :D

[Andrew]

Im siding with Occam on this one too. Regarding the calculation, I dont believe fret spacing comes into it at all. Mike's math is bang I think. Im not a mathamagician, but thats elementary school stuff.

Regarding SS frets, I do like them.. For electrics at least. With a mirror polish they play very slick, and stay that way a long time. And the tone gets a bit more bite to it. But nickel silver is definitely not obsolete.

[eliya]

That is incorrect. I'm not a mathematician either (because I don't have a PhD in math), but I did get a bachelors in math recently.

Mike's math is right if he collects points at even intervals. For instance, look at the sequence 2, 4, 6, 8. The average 5, right? and you can find it just by adding 8 and 2 and dividing by 2, or adding them all up and dividing by 4. Notice how these numbers are evenly spaced. Now let's pick something else between 2 and 8, for instance, let's pick 3 and 5, so now the sequence is 2, 3, 5, 8. The average of these four numbers is 4.5, however, if you just add 2 and 8 and divide by 2, you get 5. 

[Andrew]

I just typed up a very cogent rebuttal, and then IOS 8 froze and ate it. Now Im grumpy.

Suffice to say that it doesnt matter if the numbers are odd or even, or what the numbers of spaces is, we only need to know the total length of the total number of frets, which can be reached by multiplying the mean of the first fret and the last fret by the total number of frets. Period. (Decimal?)

I see what you saying though. I had to read it three times but I understand it now. But Id bet cold cash my calipers will agree with Mike's method every time, with a small margin of error for neck edges that arent dead straight. I cant explain it much better than that lol. Maybe Im still theoretically wrong, but it doesnt matter, cuz I have calipers and a calculator with an addition function :D.

[eliya]

My point is that mathematically, you're only guaranteed to get the correct average by adding the first and last fret and dividing by 2, only if the frets were spaced equally apart. If they're not, it's possible that this method will work, probably with some error, but it's not guaranteed to work.

Suppose the first fret is of length x, and to get the length of the next fret you add a fixed length to the previous fret. In other words, first fret is of length x, second is of length x + y, third is of length x + 2y, and so on up to x + 23y. This is the case where the frets were spaced apart equally. Now let's calculate the average by adding all frets length and dividing by 24, vs. adding the length of the first and last and dividing by 2.

(24x + 276y)/24 = x + 11.5y is the average fret length.

Now by using just the first and last fret:

(2x + 23y)/2 = x + 11.5y

Paul's method for estimating how much fretwire is needed is smart.

[Andrew]

Im not sure if I should feel humbled or dubious.

[Paul Verticchio]

Without perpetuating academics, here's how I calculate 'needed fret wire for an unbound neck"

Measure the widest part of the neck. Add 1/2" to that measurement. I like around .25" of overhang on each side of the FB to assist the fret ends in seating correctly & staying put.

Count the number of fret slots.

Multiply the number of fret slots by the "width + .5" of the fret".

Take the total and order what I need + 2 ft.

What's left over gets a tape "tag" wrapped around it that gives the source & stock # of the wire and then it goes into the random fret wire tube for future use.

Fret wire is cheap and having too much is better than having a quarter inch too little.  :)

[Andrew]

I included that method in the post that my tablet boogered up, and my pea sized brane forgoteded it. Thats basically the way to go. Except I just order the extra "4" feet. Same idea though.

[Murray MacLeod]

What matters is how many frets there are below the halfway point on the board, and how many there are above the halfway point.

In this case there are 8 frets below the halfway point and 16 frets above the halfway point. 

A moment's thought suffices to show that the average length of the frets is going to be significantly  greater than simply taking the sum of the first and last frets and dividing by 2.