# Professional Surveyor Magazine March 2008 Supplemental

## We have a problem with Problem 154

The drawing in the magazine shows the tank length as 30' but the solution on our website uses a tank of 25'. See below for a lively discussion about this accidental discrepancy as well as a proper solution.

Professional Surveyor Magazine, February 2008, Vol. 28 No. 2
Solution to Problem Corner, p. 76, Problem 154, submitted by Dan Martinez, Alexandria, VA

Approach and Solution: The area of a semicircle as a function of its diameter length (ft) is calculated, the result is doubled to account for the entire circle, and that result is multiplied by 30 to calculate the tank's volume (ft³). Function ρ presents the general solution (height (ft) of dipstick level mark) based on volume (gal) of liquid in the tank and an approximating value (ft). Respective dipstick level marks (ft) for 1000, 2000, 3000 and 4000 gallons are 1.29, 2.12, 2.87, and 3.62 feet. These results differ from those in the published solution document.

This exercise was prepared using Mathematica 6.

Plot[√3²−(x−3)², {x,0,6}, AspectRatio→Automatic, AxesLabel→{"Vertical Diameter (ft)", "Radius (ft)"}, ImageSize→300] ι[λ_]:=NIntegrate[√3²−(x−3)², {x,0,λ}] (* area (ft²) of semicircle of radius 3 feet based on distance (λ, ft) along vertical diameter *)

{π 3²,ι×2} // N (* ideal v. approximate area (ft³) of the tank's end *)

{28.27433, 28.27433}

Plot[ι[j]×2, Pj,0,6], AxesLabel→{"Distance (ft) Along Tank's Vertical Diameter", "Area (ft²) Of Tank's End"}, ImageSize→400] Integrate[√3²−(x−3)², {x,0,6}]×2×30×7.48052 (* total volume (gal) of liquid in the tank *)

6345.202

Plot[ι[j]×2×30×7.48052, {j,0,6}, AxesLabel→["Distance (ft) Along Tank's Vertical Diameter", "Volume (gal)"], ImageSize→400] σ[λ_]:=NIntegrate[√3²−(x−3)²×2×30×7.48052, {x,0,λ}] (* volume (gal) of liquid in tank based on distance (λ, ft) *)

Off[NIntegrate::"nlim"]

ρ[gal_,v_]:=Round[FindRoot[σ[λ]=gal, {λ,v}][1,2],0.01] (* dipstick level mark (ft) based on volume (gal) of liquid in tank and an approximating value (ft) *)

ρ[1000,0.8] (* distance (ft) from bottom of tank to dipstick level mark with 1000 gallons in the tank *)

1.29

ρ[2000,1.6] (* distance (ft) from bottom of tank to dipstick level mark with 2000 gallons in the tank, etc. *)

2.12

ρ[3000,2]

2.87

ρ[4000,3]

3.62

### Check of foregoing results

Map[σ, {1.29, 2.12, 2.87, 3.62}]

{1001.696, 2004.905, 2997.611, 4001.445}

### Check of volume at 3 ft dipstick level mark (should be half of the tank's volume)

σ − 6345.202 ÷ 2 (* difference between calculated and ideal volumes *)

−0.0001986007

### Check of volume at 6 ft dipstick level mark (should be all of the tank's volume)

σ − Integrate[√3²−(x−3)², {x,0,6}]×2×30×7.48052 (* difference between calculated and ideal volumes *)

1.527951×10−10

Dan,

Well, by golly, you caught me in a big mistake!

The drawing in the magazine shows a tank of length 30'. My solution used a tank of 25'! I probably changed one and not the other during my final edit.

I like your solution because it's direct. I've been trying to stay away from integrals and differentials. These problems are supposed to be able to be solved with trigonometry.

I also used 7.48 gallons per cubic foot instead of 7.48052, too.

Try your solution with a tank length of 25' and see what you get. I have no one to check my calculations except the readers. And you guys never let me get away with anything!

Thanks for your interest in the Problem Corner.

—Dave Lindell

Dear Mr Lindell: