# Equation to find countersink angle

#### wilster98

##### New member
Hello All, I am wondering if there is a way to find the angle of a countersink using the top and bottom diameters of the hole and the thickness of the material? While researching this problem in a machining forum, some one said I you have those 3 things "you can use trig to sort it out". It was an old post and my response asking for more information went un-answered. If it is indeed possible to get the angle using math, I would like to make a form in Excel for an operator to verify the angle of countersinks.
Any help would be greatly appreciated.

Thanks in advance
Willy

#### mmm4444bot

##### Super Moderator
Staff member
Code:
                   D/2
_______          _______
\     |          |     /
\    |          |    /
\   |        T | x°/
\  |          |  /
\ |          | / 90°-x°
\|          |/
Let D be the difference between the two diameters.

Let T be the thickness.

Then angle xº is arctan(D/[2·T]).

Use a calculator set to degree mode.

For example, if the diameters differ by 3/8" and the thickness is 9/16", then

D = 0.3750

T = 0.5625 so 2·T = 1.1250

arctan(0.3750/1.1250) = 18.43°

If you get 0.32175, instead, then the calculator is set to radian mode. You can multiply radians by 57.2958, to convert to degrees.

0.32175 · 57.2958 = 18.43

If you don't have a calculator with arctan, then google: arctan(0.3750/1.1250) in degrees

#### PointBlank

##### New member
Code:
                   D/2
_______          _______
\     |          |     /
\    |          |    /
\   |        T | x°/
\  |          |  /
\ |          | / 90°-x°
\|          |/
Let D be the difference between the two diameters.

Let T be the thickness.

Then angle xº is arctan(D/[2·T]).

Use a calculator set to degree mode.

For example, if the diameters differ by 3/8" and the thickness is 9/16", then

D = 0.3750

T = 0.5625 so 2·T = 1.1250

arctan(0.3750/1.1250) = 18.43°

If you get 0.32175, instead, then the calculator is set to radian mode. You can multiply radians by 57.2958, to convert to degrees.

0.32175 · 57.2958 = 18.43

If you don't have a calculator with arctan, then google: arctan(0.3750/1.1250) in degrees

is that d/2 before so if your original diameter was .75 then you divided it by to create D so its actually the ARCtanx(Radius/(2xThickness))

#### mmm4444bot

##### Super Moderator
Staff member
… is that [D/2] before so if your original diameter was .75 then you divided it by  to create D …
Hello PointBlank. No. I'd defined symbol D at the very beginning to represent the difference between the top diameter and the bottom diameter.

In my diagram, the bottom diameter is represented by the horizontal distance between the two right triangles. The combined length of the two horizontal legs of those triangles represents the difference between the two diameters. Due to symmetry, the two triangles are identical; therefore, the horizontal legs have equal length. In other words, the measure of each horizontal leg may be expressed as half the diameters' difference (D/2).

When you say, "original diameter", you're referring to the bottom diameter. Correct?

If your bottom diameter is 0.75 units, what is your top diameter?

PS: As the tangent of angle 'x' may be expressed as the ratio of sides 'Opposite/Adjacent', I could have also written the arctangent using the same ratio. That is, I could have written the ratio as [D/2]/[T] instead of D/[2T] because those are the same ratio:

arctangent([Opposite]/[Adjacent])

arctangent([D/2]/[T]) 