Shimming the Kahler Spyder for the HM Strat 17" radius.

[rickoshay]

By all means, double check my math.  Short version - I am shimming the a and d strings 0.3mm, and the E strings 0.6mm.  Below is why.  WARNING!  Math and trig content.

(originally posted in a math thread on AR!5.com)

"If you mean the distance D in the image, then the formula is R*(1-cos(a))



If you can't see it, right click and select 'open in another tab'



Mike"  imgur.com/cud9vZ4




Thanks. Solving for D for various angles will eventually get me the info I need. A guitar has 6 strings, so the point of tangency is between the 3rd (D) and 4th (G) strings. I have the necessary angles, and I believe radius as well. I have a 17 inch radius fingerboard, but a Kahler Spyder bridge which is a 14"  15" radius. {Actually 15 inches according to Kahler]  By working this problem for both a 17" and 14" 15" radius for each bridge saddle, I should be able to calculate the shim thickness needed.



Thanks again. Trig was a while ago for me.

[rickoshay]

So we have a bridge string spacing of 2.1 inches which gives us an arc length of 2.1003, which works out to an angle for the whole bridge of 16.045 degrees when the radius is 15 inches. 15 inches was the stock fretboard radius the Kahler Spyder bridge was built with.

So, per the diagram, we have the angle to get the included angle 8.0225 degrees - the cosine of which is 0.99021334. So the formula given calculates that the distance to a tangent line, raw, is 0.1467999 inches, for the high and low E strings. We will have to adjust that a little, more on that later.

We have 6 strings, equally spaced, which divides our total bridge arc into 5 equal arcs. We now need to calculate the included angle for the A and B strings, which will be half of 3/5ths of that of the total bridge, or 4.8135 degrees/

Our formula cranks out 0.0529032 inches, raw.

BUT our center two strings are also a distance form the tangent. However, we need to "Zero" our saddles heights by subtracting this distance from the other calculations! The distance from the tangent involves a central angle of half of 1/5 of 16.045 degrees, which is an easy to figure 0.0529032 degrees. The formula gives us a distance to the tangent of : 0.00000645 inches.

So our actual saddle height for the low and high E strings is 0.14679345 inches.

The A and D strings work out to 0.05289675. The effect of zeroing out for the center saddle height is very small. Let's see how things work out for a 17 inch radius!

[rickoshay]

ON a 17 inch radius, the total bridge spans an arc of 2.10008100814845 - slightly less that we get with the smaller radius. Our circumference is now 53.4071 inches. So the total angle from end to end of the center of the outer strings is (2.10008100814845/53.4071) * 360 = 14.15596733268502 degrees, half of which is 7.077983666342509 degrees.

So our raw figure for the outside E strings distance to tangent works out to 0.12955088 inches, slightly less than the 14 inch radius.

For the A and B strings, the raw figure calculates out to 0.04667622 inches.

The correction factor that we have to subtract from these, the tangent distance to the D and G strings, calculates as: 0.0051884 inches.

So our adjusted distance is:

E strings: 0.12436248 inches.

A/B strings: 0.04148782

Stay tuned for our exciting conclusion in Part III!

[rickoshay]

For the 15" - r actual saddle height (distance to tangent line) for the low and high E strings is 0.14679345 inches.

The A and D strings work out to 0.05289675.

For the 17" - E strings: 0.12436248 inches.

A/B strings: 0.04148782

We zero out in both cases because the bridge raises and lowers on two pivots, so we are setting "par" to be the height of the center two saddles in both cases, as we are not shimming them. We are changing the other saddles heights relative to them.

So what does the distance come out to be?

E strings : 0.14679345 - 0.12436248 = 0.02243097 inch. 0.569746638 mm We will need shim stock as close to that as I can get.

A/D strings: 0.05289675 - 0.04148782 = 0.01140893 inch. 0.28978682199999994 mm. If I can find 0.3 mm shim stock I will just double the shims under the outside saddles.

Factory Floyd Rose shims are coincidentally 0.15mm. I could stack those up - if I could find them.

[rickoshay]

I can't find shims so I am having to make them out of a cheap set of feeler gauges from an auto parts store. Going to put 0.3 mm under the A and B, and 0.6 mm under the Es. This will slightly flatten the bridge compared to a 17" radius, but as the strings are slightly above the neck,. inscribing a slightly larger arc should be OK. In any case, I will be closer than the factory setup is!

[The Guitar]

Neat! Might give this a shot on the next string change.

[The Guitar]

Jun 24, 2021 21:22:04 GMT -5 rickoshay said:

I can't find shims so I am having to make them out of a cheap set of feeler gauges from an auto parts store. Going to put 0.3 mm under the A and B, and 0.6 mm under the Es. This will slightly flatten the bridge compared to a 17" radius, but as the strings are slightly above the neck,. inscribing a slightly larger arc should be OK. In any case, I will be closer than the factory setup is!

Any results yet?