if im building a bass guitar with a thru-neck style with a bridge that is front loading am i supposed to angle the neck any? and what angle do i use for the head of the guitar in relation to the neck.
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mikeyswood said:Angle = arctan((distance between top of bridge and top of fingerboard) / (distance between end of fingerboard and bridge))
don't get it :/
out of my league with the math; i just make a drawing for each style of bridge![]()
Not the nut. The end of the fingerboard. This calculation if to find the cant of the neck. If you measure to the nut then this equation is useless.
But does one degree make a difference?Easy trig. Imagine a flat plane, with your bridge and nut sitting on it. Let's say the bridge saddles are .5" higher than the nut. Then you slide bridge away from the nut a distance that corresponds to your scale length; say 34" (34" would actually be the length of the hypotenuse, but it will be close enough.) Now you have two sides of a right triangle, so you can figure out the angle at the nut. From the nut, the bridge height is the Opposite, and the length from the nut is the Adjacent, so using your trusty mnemonic SOHCAHTOA, you know that this is a tangent function. You're trying to get an angle though, not the length of a side, so you use the inverse tangent (or arctan, or tan^-1, they mean the same thing.) So divide .5 by 34 on your scientific calculator and hit the tan^-1 key, and you get .84252..... about one degree.
When you change the neck angle though, you change the height of the bridge from the nut, which changes the angle...this could be a calculus problem, and I suck at that. The trig solution is close enough.
so your finding the angle to cut in the body portion that gives you perfect action height on the last fret; and the 'string' would represent the longest line of the triangle?
those arnt techy terms; but im tired and already over my head :/
Not to the nut; to the end of the fingerboard. This calculation is to find the cant of the neck. If you measure to the nut this equation is useless.