Calculating Linear Feet of Wood Needed to Form Circle

[Onlinebear]

I create striped boards out of different woods, then cut the boards into small "wedges" to glue together into a circle/column. All that to turn the column into ornaments/containers/wine stoppers. I'm looking for Excel formulas to calculate how many linear inches of board I need to create the circle/columns. Here is what I know to start the problem:

Number of Wedges: 16
Desired Outside Diameter of Circle/Column: 2.0 inches
Desired Inside Diameter of Circle/Column: 0.5 inches
Width of Sawblade: 0.10 inches
Outside width of Wedge: 0.353 inches
Inside Width of Wedge: 0.088 inches

First I cut one segment width from the end of the board to set the cutting angle.
Second I flip the board over and cut the first wedge. Then flipping the board again to cut second wedge... and so on. This flipping wastes only the width of the saw blade for each wedge.

I've been using this formula, but I end up using significantly more linear feet of wood. Where did I go wrong

=((DesiredNumberOfSegments/2)*OutsideSegmentLength)+((DesiredNumberOfSegments/2)*InsideSegmentLength)+((DesiredNumberOfSegments+2)*BladeWidth)+(2*OutsideSegmentLength)

. . .

 

[Dr.Peterson]

I create striped boards out of different woods, then cut the boards into small "wedges" to glue together into a circle/column. All that to turn the column into ornaments/containers/wine stoppers. I'm looking for Excel formulas to calculate how many linear inches of board I need to create the circle/columns. Here is what I know to start the problem:

Number of Wedges: 16
Desired Outside Diameter of Circle/Column: 2.0 inches
Desired Inside Diameter of Circle/Column: 0.5 inches
Width of Sawblade: 0.10 inches
Outside width of Wedge: 0.353 inches
Inside Width of Wedge: 0.088 inches

First I cut one segment width from the end of the board to set the cutting angle.
Second I flip the board over and cut the first wedge. Then flipping the board again to cut second wedge... and so on. This flipping wastes only the width of the saw blade for each wedge.

I've been using this formula, but I end up using significantly more linear feet of wood. Where did I go wrong

=((DesiredNumberOfSegments/2)*OutsideSegmentLength)+((DesiredNumberOfSegments/2)*InsideSegmentLength)+((DesiredNumberOfSegments+2)*BladeWidth)+(2*OutsideSegmentLength)

First, let's shorten the variable names to save me a little writing:

N = desired number of segments
A = outside segment length
B = inside segment length
W = blade width

So the formula is: (N/2)(A) + (N/2)(B) + (N + 2)(W) + 2(A)

Now, here's how I picture you cutting N wedges (and a little scrap) from one board:

[FONT=courier new]    B W    A    W B W    A
+--+-+ +-------+ [/FONT][FONT=courier new][FONT=courier new]+-+[FONT=courier new] +-------+
[/FONT][/FONT]|:/   \ \     / /   \ \     /|
|/     \ \   / /     \ \   /:|
+-------+ +-+[FONT=courier new] +-------+ +-+--+
    A    W B W    A    W B
[/FONT][/FONT]

Here N = 4, and A, B, and W are as shown. Your formula gives 2A + 2B + 6W + 2A = 4A + 2B + 6W; this is 3W + 2A more than my sketch requires, so I'm probably not quite seeing it as you do. The only thing I see wrong here, besides that, is that the width W will actually be a little more than the width of the blade itself, because the cuts are at an angle. It would actually be the width of the blade divided by the cosine of the angle from vertical. (I don't know whether the kerf might actually be more than the width of the blade, which could be another issue.) But I don't think that would have a big effect. On the other hand, even small errors of measurement could add up.

I have a few questions for you:

1. Are you confident of your calculations of the widths of the wedge? Do they actually give you the diameters you want? If so, then that can't be the issue. But you didn't show how you calculated those.
2. Do you think you are actually carrying out the plan, and not possibly making cuts farther apart than you meant to? (Just asking.)
3. Can you tell us, in a specific example (such as the one you gave), what length you got from the formula, and what actual length you needed? That would give us a better idea of what kind of error we are looking for.

 

[Onlinebear]

Some Additional Info

Dr Peterson... thanks for your response!.

Your illustration is very close. The 2 extra blade widths in (N+2)(W) are one at each end of the board (Probably should be +1W). I also add 2A for the scrap wood used at each end of the board.

I measure and use the kerf width in the formula. This has been maybe 0.03-0.05 inches wider than the blade.

The Cosine reference went over my head... sorry. Should W = (KerfWidth/Cosine of CutAngle)

Does the formula assume a 1 inch board thickness? My boards are usually 0.6-0.8 inches thick.

#1 Here are my wedge width formulas:
Outside Radius = DesiredOutsideDiameter/2
Inside Radius = DesiredInsideDiameter/2
Segment Angle = 360/N
Cut Angle = SegmentAngle/2
A = 2*DesiredOutsideRadius*TAN(RADIANS(SegmentAngle/2))
B = 2*DesiredInsideRadius*TAN(RADIANS(SegmentAngle/2))

#2 I'm pretty confident in how I carry out the plan. If I have width variance to all segments, I'll update Excel, but still get several extra wedges than what my formula reports. If one wedge is off, I don't get a clean circle.

#3 I do not have an exact example. I can say I ALWAYS get more wedges than the formula reports. I'm away from my shop for a few weeks or I'd cut one and post the results.