CogoKit

Area & partitioning

Partitioning a parcel by a cut line of given direction

Three acres are to be severed from an eight-acre tract by a line parallel to the frontage. Set up the cut area as a quadratic in the perpendicular offset, solve it, and stake the two new corners.

Analyze· about 32 minutes by hand· 5 steps

Given

  • Parcel: TRACT 7, four corners, US survey feet, northing first.
  • A N 4700.00 E 6200.00
  • B N 5100.00 E 6500.00
  • C N 4820.00 E 7040.00
  • D N 4280.00 E 6760.00
  • Boundary courses: A–B N 36°52′12″ E 500.00 ft; B–C S 62°35′33″ E 608.28 ft; C–D S 27°24′27″ W 608.28 ft; D–A N 53°07′48″ W 700.00 ft.
  • The owner wishes to sever exactly 3.000 acres from the A–B end of the tract.
  • The cut line must run parallel to the A–B frontage. Call its intersection with side A–D point A′ and its intersection with side B–C point B′.

Required

  • The perpendicular offset h of the cut line from A–B.
  • Coordinates of A′ and B′, and the length and bearing of the new line A′–B′.
  • The area of the remainder.

Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.

Worked solution

Total area first, so the parts can be checked against a whole

Run the shoelace sum on A, B, C, D. Twice the area is −720,000.00 ft², so the tract is 360,000 ft², or 8.264 acres. Doing this before the partition is not optional bookkeeping: the two pieces have to add back to it, and without the total there is nothing to add back to.

Severing 3.000 acres leaves 5.264 acres. Write that number down now — it is the target for the final check, computed independently of everything that follows.

2A = −720,000.00 ft² → A = 360,000 ft² 360,000 / 43,560 = 8.264463 ac remainder to expect = 360,000 − 3.000 × 43,560 = 360,000 − 130,680 = 229,320 ft² = 5.264 ac

Resolve the two flanking sides along and across the frontage

A cut line parallel to A–B is fixed by one number: its perpendicular distance h from A–B. So resolve each of the two sides the cut line will cross into a component along A–B and a component perpendicular to it. Along-A–B is the unit vector (cos 36°52′12″, sin 36°52′12″) = (0.8000, 0.6000) in northing and easting; perpendicular is (−0.6000, 0.8000).

Side A–D is the simple case: its bearing S 53°07′48″ E is exactly 90° from A–B, so A–D is purely perpendicular. Point A′ therefore sits a distance h from A measured straight down that side.

Side B–C is not perpendicular. Resolving its vector (−280.00, +540.00) gives +600.00 ft across the frontage and +100.00 ft along it. So as the cut line moves h across the tract, B′ also slides 100/600 of h further along, and the cut line grows longer as it moves. That is what makes the area quadratic in h rather than linear.

A–B unit vector (ΔN, ΔE) = (0.8000, 0.6000); perpendicular unit = (−0.6000, 0.8000) A–D vector = (−420.00, +560.00): along = −336.00 + 336.00 = 0.00, across = +252.00 + 448.00 = +700.00 B–C vector = (−280.00, +540.00): along = −224.00 + 324.00 = +100.00, across = +168.00 + 432.00 = +600.00 so length of the cut line at offset h is A′B′ = 500.00 + (100.00 / 600.00)·h = 500.00 + h/6

Write the cut area as a function of h and solve

The severed piece A–B–B′–A′ is a trapezoid: two parallel sides of length 500.00 and 500.00 + h/6, separated by h. Its area is the mean of the parallel sides times the separation.

Setting that equal to 130,680 ft² gives a quadratic in h. Multiply through by 12 to clear the fraction and solve with the positive root — the negative root places the cut line outside the tract on the far side of A–B and has no meaning here.

Before accepting the root, confirm the cut line stays within the sides it is supposed to cross. Side A–D runs 700.00 ft across and side B–C runs 600.00 ft across, so h must not exceed 600.00 ft or the cut line would run past corner C and the trapezoid model would collapse. h = 250.87 ft is comfortably inside that.

A_cut = ½ · ( 500.00 + 500.00 + h/6 ) · h = 500h + h²/12 500h + h²/12 = 3.000 × 43,560 = 130,680 h² + 6000h − 1,568,160 = 0 h = ( −6000 + √( 6000² + 4 × 1,568,160 ) ) / 2 = ( −6000 + √42,272,640 ) / 2 √42,272,640 = 6501.7413 h = ( −6000 + 6501.7413 ) / 2 = 250.8707 → h = 250.87 ft valid: 250.87 < 600.00, so the cut line crosses B–C and not the far corner

Fix the two new corners on the ground

A′ lies on side A–D. Since that side is perpendicular to the frontage, the distance from A to A′ is h itself, 250.87 ft on bearing S 53°07′48″ E.

B′ lies on side B–C at the fraction h/600 of that side's length, because B–C covers 600.00 ft of perpendicular offset over its full 608.28 ft. That is 250.87/600.00 × 608.28 = 254.33 ft from B on bearing S 62°35′33″ E.

Compute the coordinates from those two calls, then inverse A′ to B′ to get the cut line as it will be recorded.

New corners and the calls that set them
PointSet fromBearingDistance (ft)Northing (ft)Easting (ft)
A′AS 53°07′48″ E250.874549.486400.70
B′BS 62°35′33″ E254.334982.936725.78
A′–B′cut lineN 36°52′09″ E541.81
A′–Dremainder sideS 53°07′48″ E449.13
B′–Cremainder sideS 62°35′33″ E353.95

Areas of the two pieces

The severed piece A–B–B′–A′ is 130,680 ft², or exactly 3.000 acres, by construction. The remainder A′–B′–C–D is the difference, 229,320 ft² or 5.264 acres.

Recomputing the pieces from the staked coordinates gives 130,678.7 ft² and 229,320.2 ft². The 1.3 ft² shortfall in the severed piece is not an error in the partition — it is the consequence of rounding A′ and B′ to 0.01 ft, which shifts each of them a few thousandths of a foot off the mathematical side line. On a 500 ft frontage a 0.004 ft shift is worth about 2 ft², and 2 ft² is 0.00005 acre. The deed should call 3.000 acres.

Note also that the inversed cut line bearing, N 36°52′09″ E, differs by three seconds from the A–B bearing it is meant to parallel. That is the same rounding, seen from the other end. State the cut line as parallel to A–B in the description, and let the coordinates be what they are.

A_cut = ½ (500.00 + 541.81) × 250.87 = ½ × 1041.81 × 250.87 = 130,679.4 ft² (from rounded values) A_cut exact from the quadratic = 130,680 ft² = 3.000 ac A_remainder = 360,000 − 130,680 = 229,320 ft² = 5.264 ac

Answer

  • The cut line lies h = 250.87 ft perpendicularly from the A–B frontage.
  • A′ = N 4549.48, E 6400.70, being 250.87 ft from A on S 53°07′48″ E along side A–D.
  • B′ = N 4982.93, E 6725.78, being 254.33 ft from B on S 62°35′33″ E along side B–C.
  • The new line A′–B′ is 541.81 ft long, parallel to A–B.
  • Severed parcel A–B–B′–A′ = 130,680 ft² = 3.000 acres.
  • Remainder A′–B′–C–D = 229,320 ft² = 5.264 acres.

Check

Shoelace the two pieces separately from their staked coordinates: A–B–B′–A′ returns 130,678.7 ft² and A′–B′–C–D returns 229,320.2 ft². They sum to 359,998.9 ft² against a whole tract of 360,000 ft².

The 1.1 ft² shortfall is entirely attributable to rounding the two new corners to 0.01 ft; it is 3 parts per million of the tract and 0.00003 acre. If it were tens of square feet, the quadratic root or one of the two stakeout calls would be wrong.

Independent check on the remaining side lengths: A′ must be 700.00 − 250.87 = 449.13 ft from D, and B′ must be 608.28 − 254.33 = 353.95 ft from C. Both inverse to exactly those values from the staked coordinates.

Check the trapezoid formula against the geometry a second way: the cut line grew from 500.00 ft at h = 0 to 541.81 ft at h = 250.87, a growth of 41.81 ft, and h/6 = 250.87/6 = 41.81 ft. The along-frontage resolution of side B–C is confirmed.

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