# Finding base circle with coordinates of two points of involute curve

1. Jan 22, 2014

### k.udhay

Hi,

I have an involute gear and measured co-ordinates of two arbitrarily chosen points (on the involute portion) of a tooth. Can I find out the base circle with this information? Thanks.

2. Jan 22, 2014

### Baluncore

Theoretically, yes, but only if you know two involute point locations relative to the centre of the gear.

The centre of a gear is usually hollow or inaccessible, so you will need to measure coordinates of three points, then compute a circle centre position.

The base circle is related to points on the involute by the length of base circle tangents to those points.
Because the gear contact angle is typically 15° to 20°, the two equations that must be solved involve a very long triangle. The error in your measurement is therefore doubled, (because there are two points), and then multiplied by at least 1/Sin(20°)=3. That amplifies your measurement errors by at least 6 for your base circle estimate, to which you must add your centre position error.

There are better ways of computing the base circle diameter from measurements of the pitch circle and counting the teeth. The base circle is needed to mathematically generate an involute gear tooth profile, but is irrelevant to the generation or functioning of a gear wheel.

Are you sure you want to go ahead with your error amplifying two point solution?

3. Jan 22, 2014

### k.udhay

Hi Baluncore,

I actually want to find out the pressure angle of the gear in my hand (standard or operating). Knowing the operating centre distances from the housing bores locations, I thought I should find base circle diameter to find the operating pressure angle. This was the intention of the activity.

If I can get either the pressure angle or base circle dia easier than the method I asked for, pl. share it. Kindly note that I don't have any dedicated gear parameters measuring machine. All I have is a CMM and some basic workshop tools. Thanks a ton!

P.s. - Out of curiosity, can you pl. share any equation available to find the base circle dia from coordinates of two points (WRT the base circle centre) of an involute? I tried deriving only to find that I am bad with it. Thanks again!

4. Jan 22, 2014

### Baluncore

Measure the coordinates of two points on the involute gear face, symmetrically about the pitch circle contact.
Define a straight line through the two measured points.
Compute the midpoint of the two measured points.
Define a line through the gear wheel centre and the computed midpoint.
The angle between those two defined lines is the contact angle.

The equations for computation of base circle radius from two points on the involute will have to wait until tonight.

5. Jan 23, 2014

### k.udhay

I tried this in AutoCAD with a dxf of involute generated by KISSsoft. A very good method to determine pressure angle. I think the error / approximation increases as the distance between the two points to be measured and the pitch circle contact increases?

About my originally proposed method, one of my friends has given me this hint:

Thanks for your constant help, Baluncore!

6. Jan 25, 2014

### Baluncore

Here is my algorithm for computation of the base circle radius from two points on the involute.

Given two points A(x,y), B(x,y) on the involute and the gear wheel centre C(x,y)

Reduce all coordinates to place gear wheel centre at the origin
A = A – C
B = B – C
C = C – C

Compute the radial distance to points A and B from the gear centre

Select an initial estimate for the base circle radius, R, { 0 < R < Ra }

Lengths from point on involute along tangent to the base circle contact
Sa = Sqrt( Ra^2 – R^2 )
Sb = Sqrt( Rb^2 – R^2 )

By the definition of an involute as an unwinding filament we know
arc = Sa – Sb

As a filament unwinds from the base circle, it describes an involute through points A and B
Solve for the base circle tangent contact points P and Q for points A and B respectively

The intersection of two circles, (you must select the correct point from the two available)
circle centre A, radius Sa; with centre C, radius R; gives point P
circle centre B, radius Sb; with centre C, radius R; gives point Q

We now have two independent estimates for the chord based on base circle radius, R
chord1 = Radius( P – Q ) the linear distance between the two contact points on base circle
chord2 = 2 * R * Sin( 0.5 * arc / R ) the relationship between radius, arc and the chord

A search for chord1 – chord2 = 0 is easier than a solution of equations
When these two chord values are equal, we have R, the base circle radius

I have checked that it does work for real values.

7. Feb 2, 2014

### k.udhay

Hi Baluncore,
Thanks for your detailed explanation. And sorry for my late reply. Went out of station for some time with limited internet access.