Layout and shop math
Area calculator for bar stock, plate blanks and chips
A 1.000 in round bar has 0.7854 in² (506.71 mm²) of cross section; mill a 0.500 in flat on it and 0.7628 in² (492.10 mm²) is left.
Technical review pending. Formulas and values are cited. This notice comes down after a machinist review.
Cross-section area
in²
A = (pi / 4) x D^2 - As = (3.1416 / 4) x 1^2 - 0.022647 = 0.7628 in²
492.10 mm², 4.92 cm². theta = 2 x asin(c / (2 x R)) = 2 x asin(0.5 / (2 x 0.5)) = 1.047198 rad (60.0000°). As = (R^2 / 2) x (theta - sin(theta)) = (0.5^2 / 2) x (1.047198 - sin(1.047198)) = 0.0226 in² removed, 2.88% of the section. Flat depth h = R - sqrt(R^2 - c^2 / 4) = 0.5 - sqrt(0.5^2 - 0.5^2 / 4) = 0.0670 in.
Multiply by cutting speed or feed rate for metal removal rate; the metal removal rate calculator adds spindle power.
Source: Machinery's Handbook (areas of plane figures; segments of circles); plane geometry as shown. Weight uses the densities on the metal weight calculator. 1 in² = 645.16 mm² (25.4 mm per inch, exactly).
Areas are nominal geometry from the sizes entered. Real stock runs a few thousandths off nominal, and corners are rarely sharp.
How it works
| Shape | Area formula | Example |
|---|---|---|
| Round bar | A = (pi / 4) x D^2 | D 1.000 in: 0.7854 in² |
| Square bar | A = a^2 | a 1.000 in: 1.0000 in² |
| Hex bar | A = 0.866025 x AF^2 | AF 1.000 in: 0.8660 in² |
| Octagon bar | A = 0.828427 x AF^2, from 2 x (sqrt(2) - 1) | AF 1.000 in: 0.8284 in² |
| Flat bar or rectangle | A = W x T | 1.000 x 0.250 in: 0.2500 in² |
| Round tube | A = (pi / 4) x (OD^2 - ID^2); ID = OD - 2 x t | 1.000 in OD x 0.750 in ID: 0.3436 in² |
| Rectangular tube | A = W x H - (W - 2 x t) x (H - 2 x t) | 1.000 x 1.000 in, 0.065 in wall: 0.2431 in² |
| Round bar with one flat | A = (pi / 4) x D^2 - As; As = (R^2 / 2) x (theta - sin(theta)) | D 1.000 in, 0.500 in flat: 0.7628 in² |
| Plate blank with holes | A = W x L - n x (pi / 4) x d^2 | 1.000 x 1.000 in, one 0.250 in hole: 0.9509 in² |
| Chip, turning | A = ap x f | 0.100 in x 0.010 IPR: 0.0010 in² |
| Slot or side cut | A = WOC x DOC | 0.500 x 0.250 in: 0.1250 in² |
For a flat on round bar, theta = 2 x asin(c / (2 x R)) from the flat width or 2 x acos((R - h) / R) from the depth, in radians. Inch areas show 4 places, mm² and cm² 2 places.
Key formulas
A = (pi / 4) x D^2; As = (R^2 / 2) x (theta - sin(theta)); 1 in² = 645.16 mm²
| Symbol | Meaning | Unit |
|---|---|---|
| A | Cross-section area | in² or mm² |
| D, R | Bar diameter; R = D / 2 | in or mm |
| a | Side of a square bar | in or mm |
| AF | Across flats of hex or octagon bar | in or mm |
| W, T, L | Width, thickness and plate length | in or mm |
| OD, ID, t | Tube outside and inside diameter, wall | in or mm |
| H | Rectangular tube height | in or mm |
| c, h | Flat width (chord) and flat depth (segment height) | in or mm |
| theta, As | Segment angle and segment area removed by the flat | rad; in² or mm² |
| n, d | Number of holes and hole diameter | count; in or mm |
| ap, DOC | Depth of cut | in or mm |
| f | Feed per revolution | IPR or mm/rev |
| WOC, ae | Width of cut | in or mm |
| SFM | Surface feet per minute, cutting speed | ft/min |
| Vc | Cutting speed, metric | m/min |
| IPM, vf | Inches per minute, table feed rate; vf in mm/min | in/min or mm/min |
| MRR | Metal removal rate, inch | in³/min |
| Q | Metal removal rate, metric | cm³/min |
Worked example: 1.000 in round bar with a 0.500 in flat
Round bar, D = 1.000 in (R = 0.500 in), flat width c = 0.500 in.
- Full round: A = 0.785398 x 1.000^2 = 0.7854 in² (506.71 mm²).
- theta = 2 x asin(0.500 / 1.000) = 2 x 30° = 60.0000° = 1.047198 rad; sin(theta) = 0.866025.
- As = (0.500^2 / 2) x (1.047198 - 0.866025) = 0.125 x 0.181172 = 0.0226 in² (14.61 mm²).
- A = 0.785398 - 0.022647 = 0.7628 in² (492.10 mm²); the flat removes 2.88% of the section.
- Flat depth: h = 0.500 - sqrt(0.2500 - 0.0625) = 0.0670 in.
- Weight per foot in 6061 at 0.0975 lb/in³: 0.762752 x 12.000 x 0.0975 = 0.89 lb, against 0.92 lb for the full round.
Result: 0.7628 in² (492.10 mm²) left after the flat.
Worked example: steel plate blank with a bore (metric)
Carbon steel blank 200 x 150 mm, 12 mm thick, one 50 mm bore; 7.85 g/cm³ from the weight page table.
- Blank: A = 200 x 150 = 30,000.00 mm² (300.00 cm², 46.5001 in²).
- Bore: 0.785398 x 50^2 = 1,963.50 mm². Net A = 30,000 - 1,963.50 = 28,036.50 mm² (280.37 cm², 43.4567 in²).
- Volume = 28,036.50 x 12 / 1000 = 336.44 cm³.
- Weight = 336.44 x 7.85 / 1000 = 2.64 kg (5.82 lb).
- Without the bore: 30,000 x 12 / 1000 = 360.00 cm³; 2.83 kg (6.23 lb), what the metal weight calculator gives for a 200 x 150 x 12 mm plate.
Result: 28,036.50 mm² net, 2.64 kg (5.82 lb).
Short checks
- Hex 1.000 in AF: 0.866025 x 1.000^2 = 0.8660 in² (558.72 mm²); x 12.000 in = 10.392 in³, the volume in the weight page's hex example.
- Octagon 1.000 in AF: 0.8284 in² (534.47 mm²). Square 1.000 in: 1.0000 in² (645.16 mm²).
- Round tube 2.000 in OD x 1.500 in ID: 0.785398 x (4.000 - 2.250) = 1.3744 in² (886.74 mm²).
- Chip, turning: ap 0.100 in x f 0.010 IPR = 0.0010 in² (0.65 mm²); at 400 SFM, MRR = 0.0010 x 12 x 400 = 4.800 in³/min. Metric: 2 mm x 0.25 mm/rev = 0.50 mm²; at 150 m/min, Q = 0.50 x 150 = 75.00 cm³/min.
- Slot: WOC 0.500 in x DOC 0.250 in = 0.1250 in² (80.65 mm²); at 12.0 IPM, MRR = 1.500 in³/min.
Shop notes
- Hex and square bar with broken or rounded corners has a little less area than the sharp-corner formula.
- The chip cross-section ap x f is the uncut chip area for a straight turning cut; the chip that comes off is thicker and shorter than that.
- For a flat on a shaft, the depth that matters on the drawing is usually h; the calculator takes either the width or the depth.
- A plate blank's weight here uses nominal density from the weight page table; saw kerf and mill tolerance are not included.
FAQ
What is the cross-section area of 1 in round bar?
0.7854 in² (506.71 mm²): (pi / 4) x 1.000^2.
What is the area of a 1 in hex bar?
0.8660 in² (558.72 mm²) across 1.000 in flats: 0.866025 x AF^2. A 1.000 in octagon is 0.8284 in².
How much does a 0.500 in flat take off 1 in round bar?
0.0226 in², 2.88% of the section; 0.7628 in² is left, and the flat is 0.0670 in deep.
What is the chip area for a 0.100 in cut at 0.010 IPR?
0.0010 in² (0.65 mm²). Times 12 x 400 SFM, that is 4.800 in³/min of metal removal.
How do you convert in² to mm²?
Multiply by 645.16, which is 25.4 x 25.4 exactly: 0.7854 in² is 506.71 mm².
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