F-Engrave V-Carve Calculations

Within F-engrave all parts of the font characters are

converted into short line segments. To determine the

toolpath for V-Carving we step along each line segment

that defines the text. (the size of the step is

adjustable in the V-Carve Settings). At each step along

each segment the maximum radius is determined. The max radius

circle always passes through the current step point and is

centered on a the Y-axis of the local coordinate system.

The local coordinate system is centered on the current

step point and the Y-axis is generally perpendicular to

the line segment. At sharp turns from segment to segment

the local coordinate system is stepped between the

current line segment and the next line segment to avoid

cutting the corner.

converted into short line segments. To determine the

toolpath for V-Carving we step along each line segment

that defines the text. (the size of the step is

adjustable in the V-Carve Settings). At each step along

each segment the maximum radius is determined. The max radius

circle always passes through the current step point and is

centered on a the Y-axis of the local coordinate system.

The local coordinate system is centered on the current

step point and the Y-axis is generally perpendicular to

the line segment. At sharp turns from segment to segment

the local coordinate system is stepped between the

current line segment and the next line segment to avoid

cutting the corner.

The calculation is dependant on the font being defined by

loops that are created CW (clockwise). Internal

loops are defined CCW (counter clockwise). If this

convention is reversed the flip normals option in

F-Engrave will correct the calculation.

loops that are created CW (clockwise). Internal

loops are defined CCW (counter clockwise). If this

convention is reversed the flip normals option in

F-Engrave will correct the calculation.

Figure 1

We start with a couple of special cases. The first special

case is x1==x2. If x1==x2 (Figure 1) the line segment

is vertical in the local coordinate system. For this

case we simply set R to X1 (or X2) and check to make

sure the line segment exists at y=R. If the line segment

exists at y=R we save the Radius R. If all of the line

segment is above or below y=R the result is discarded.

case is x1==x2. If x1==x2 (Figure 1) the line segment

is vertical in the local coordinate system. For this

case we simply set R to X1 (or X2) and check to make

sure the line segment exists at y=R. If the line segment

exists at y=R we save the Radius R. If all of the line

segment is above or below y=R the result is discarded.

Figure 2

The second special case is y1==y2. If y1==y2 (Figure 2)

the line segment is horizontal in the local coordinate

system. For this case we check to make sure y1 (or y2)>0.

If y1 is >0 we set R to y1*1/2. We also check to make sure

the line segment exists at the x=0. If the line segment

exists at x=0 we save the Radius R. If not the result

is discarded.

the line segment is horizontal in the local coordinate

system. For this case we check to make sure y1 (or y2)>0.

If y1 is >0 we set R to y1*1/2. We also check to make sure

the line segment exists at the x=0. If the line segment

exists at x=0 we save the Radius R. If not the result

is discarded.

Figure 3

Next we move to a more general case. This case is a line segment

with a arbitrary slope (Figure 3). This case requires a bit of

algebra. we start by looking at the geometry and set up a few

governing equations. First we need the equation of a line that

passes through the two points in the line segment (eq1).

with a arbitrary slope (Figure 3). This case requires a bit of

algebra. we start by looking at the geometry and set up a few

governing equations. First we need the equation of a line that

passes through the two points in the line segment (eq1).

(eq1)

The second equation is the equation for a line that passes though

the center of our circle and is perpendicular to the first line. The

first line has a slope of m. To make the second line perpendicular

to the first line we set it's slope to -1/m. Since we know the slope

and we know the line passes through the y-axis at R (y-intercept)

we can write the equation for the second line as (eq2).

the center of our circle and is perpendicular to the first line. The

first line has a slope of m. To make the second line perpendicular

to the first line we set it's slope to -1/m. Since we know the slope

and we know the line passes through the y-axis at R (y-intercept)

we can write the equation for the second line as (eq2).

(eq2)

The third and last governing equation (eq3) is the equation for

a line circle with radius R centered at x=0, y=R. (Figure 3.)

a line circle with radius R centered at x=0, y=R. (Figure 3.)

(eq3)

Combining (eq1) and (eq2):

Solve for R:

Create a new variable Lambda to

make the equations less messy:

make the equations less messy:

(eq4)

(eq5)

Expand (eq3):

Cancel out the squared R values and

Substitute for R and y using (eq1)

and (eq5)

Substitute for R and y using (eq1)

and (eq5)

The first term is simply:

Expanding the second term:

Expanding the third term:

All three terms together expands to:

After combining the b squared terms, canceling

out the 2bmx terms and grouping we get (eq6) which

is in the form of the quadratic equation:

(the equations in red below are used in the

F-Engrave python file)

out the 2bmx terms and grouping we get (eq6) which

is in the form of the quadratic equation:

(the equations in red below are used in the

F-Engrave python file)

(eq6)

Quadratic

Equation:

Equation:

Where:

(eq4)

m = Slope of the line segment

and

b = Y intercept of line passing through the line

segment points

We know the two end points of the line segment so we

can determine the equation of a line passing through

those two point (x1,y1) and (x2,y2) using (eq1)

and

b = Y intercept of line passing through the line

segment points

We know the two end points of the line segment so we

can determine the equation of a line passing through

those two point (x1,y1) and (x2,y2) using (eq1)

(eq1)

Substitute the 1st point coordinates into (eq1):

Solve for b:

(eq7)

Substitute the 1st point coordinates into (eq1):

(eq8)

Substitute the equation solved for b into (eq8)

and solve for m:

and solve for m:

(eq9)

Similar to the first two special cases we need to check if

the line segment exists at the intersection point and save

the Radius R if it does. We do this by checking

that min(x1,x2) < x < max(x1,x2).

the line segment exists at the intersection point and save

the Radius R if it does. We do this by checking

that min(x1,x2) < x < max(x1,x2).

Figure 5

Figure 4

The last two cases are when the line segment end points limit

the size of the max circle (Figure 4 and Figure 5). For this

case (eq3) from above is used again. This equation is performed

for both point (x1,y1) and (x2,y2).

the size of the max circle (Figure 4 and Figure 5). For this

case (eq3) from above is used again. This equation is performed

for both point (x1,y1) and (x2,y2).

(eq3)

And for (x2,y2):

After these calculations are performed for every segment

the minimum R value has been determined and its center and

radius are saved for generating the tool path. The Radius

determines the depth of the cut as a function of the

v-cutter angle and max diameter.

the minimum R value has been determined and its center and

radius are saved for generating the tool path. The Radius

determines the depth of the cut as a function of the

v-cutter angle and max diameter.