T-O-M position?

[gilsolomon]

Hi Guys,

I'm planning my next guitar (making CAD files since I'm laser-cutting plexiglass templates), It's going to be a hollowbody guitar. The body’s core will be made from American Swamp Ash, the top will be made from Brazilian flamed Imbuia, the neck will be made from African Mahogany and the fingerboard and binding will be made from Bloodwood (had a question about it a few days ago too ).

So on to my question... The bridge I'll be using is a Gotoh TOM Bridge.

I've been messing around with Stewmac's fret position calculator and couldn't understand something.

The distance they give you there is from the middle of the nut's face to the middle of the bridge's post itself of the a parallel position on the middle line of the body?

What is the recommended angle for the bridge? I saw some people put it 5 degrees some at 8... what would you say?

Thanks

Edited January 11, 2014 by gilsolomon

[guitar2005]

The distance is from the edge of the nut i.e. the start of the fretboard to the full scale length.

Their calculator includes the treble and bass post positioning.

[gilsolomon]

Yeah I understood that one... I meant the other end....
The distance is measured to the posts themselfs and not the a parallel point on the center line right?

[HuntinDoug]

Commonly, The measurement is from the leading edge of the nut, to the center of the "throw" on the high E saddle. Most TOM bridges have the studs centered in the bridge, so you can use that as well. Keep in mind, many guitar manufacturers add 1/8" to the scale length on the low E side. This makes the bridge look crooked. It is done this way for intonation purposes so there is sufficient throw:

CAD it out so the bridge is parallel with the leading edge of the nut, and then offset the low E side by adding .125" to the scale length measurement.

[gilsolomon]

Commonly, The measurement is from the leading edge of the nut, to the center of the "throw" on the high E saddle. Most TOM bridges have the studs centered in the bridge, so you can use that as well. Keep in mind, many guitar manufacturers add 1/8" to the scale length on the low E side. This makes the bridge look crooked. It is done this way for intonation purposes so there is sufficient throw:

CAD it out so the bridge is parallel with the leading edge of the nut, and then offset the low E side by adding .125" to the scale length measurement.

Thanks !

And where do I start to measure on the nut-fretboard meeting point? the center line fo the neck or one of the fretboard edges?

My current mesurement is from the treble side post to the meeting point of the nut and fretboard on the center line (The green line in the image below). Is that the correct way to measure?

Edited January 11, 2014 by gilsolomon

[LightninMike]

You would measure it from where each E string would fall on the nut, not a centerline measurement

[HuntinDoug]

If you want to use the center line, thats fine. But, you need to stay on the center line. Make your measurement from the leading edge of the nut, to the center of the bridge. Place your bridge vector perpendicular to the center line. Using the center point of the high E stud as a fulcrum point. Rotate the the low E bridge stud counter clockwise until you add .125" to the length on the low E side...

Although, It looks like the bridge vector is already adjusted in the pic. If that is the case, you can use a centerline that is your scale length + half of the offset:

25.5" + (.125/2) = 25.5625

[gilsolomon]

If you want to use the center line, thats fine. But, you need to stay on the center line. Make your measurement from the leading edge of the nut, to the center of the bridge. Place your bridge vector perpendicular to the center line. Using the center point of the high E stud as a fulcrum point. Rotate the the low E bridge stud counter clockwise until you add .125" to the length on the low E side...

Although, It looks like the bridge vector is already adjusted in the pic. If that is the case, you can use a centerline that is your scale length + half of the offset:

25.5" + (.125/2) = 25.5625

Thanks a lot for the elaborate explanation ! I finally understood it !