Issue 016 - Water infrastructure - Construction quantity takeoff
How big is a Lake Powell boat-ramp extension?
AP reports that Lake Powell is about 23% full and roughly 33 feet above the minimum level needed for hydropower generation. Low water has already forced marinas to extend, relocate, or abandon boat ramps. Convert another 33-foot drop into ramp length, concrete, truckloads, and cost.
The problem
Estimate the length of additional boat ramp needed if Lake Powell falls another 33 vertical feet.
Then estimate the volume and mass of concrete required to build that extension for one typical two-lane ramp. Convert the result into concrete-truck loads, construction cost, and ramp length compared with a football field.
Does adapting one boat ramp to a 33-foot decline resemble an ordinary paving project, or a substantial civil-engineering project?
Because Fermi problems target an order of magnitude, I normally use no more than two significant digits and write most calculations in scientific notation; the Fermi reference explains both conventions.
Before checking sources
Matt's first pass
If I am only considering a single ramp, and only considering the additional drop of 33 feet, or about 11 meters, then my assumptions are:
- boat ramps are built at approximately a 30-degree angle from horizontal;
- I can use 3-4-5 triangles to approximate the lengths involved;
- a two-lane ramp is about 7 meters wide and about 0.5 meters thick;
- concrete density is about 8 x 10^3 kg/m3.
I complicated the geometry because I am trying to do these problems with mental math rather than calculators, so I avoided using a trigonometric function like sine. Using my triangle setup, I got a ramp length of about 19 meters, then added about 10% for the angle difference.
vertical drop ~= 11 m
estimated ramp length ~= 21 m
concrete volume ~= length x width x thickness
~= (21 m) x (7 m) x (0.5 m)
~= 7.3 x 10^1 m3
concrete mass ~= volume x density
~= (7.3 x 10^1 m3) x (8 x 10^3 kg/m3)
~= 6 x 10^5 kgThat is about 600 metric tons of concrete. That sounds like it is probably in the range of a substantial civil-engineering project.
Calibration Score
Matt's Calibration Score: 40 / 100
Higher is better: earn points for accurate pegs, sound models, correct math, and a result close to the sourced answer. The image shows percent full of it: 100 minus the Calibration Score.
Pegs: 0/30. Ramp slope, concrete density, and slab-thickness pegs were meaningfully off.
Model: 15/30. The geometry and quantity-takeoff model was right in spirit but overcomplicated.
Math: 5/10. The truckload part was missed and some unit pegs were shaky.
Result: 20/30. Canceling errors kept the answer within an order of magnitude.
Grounding facts
A 33-foot vertical decline sounds modest until you divide by a shallow trailer-friendly slope. At a 12% to 15% grade, every vertical foot of lake decline requires about 7 to 8 horizontal feet of extra ramp.
That means each additional 10 feet of vertical lake decline can require about 70 to 80 feet of ramp. A 33-foot drop can consume most of a football field before you have even counted staging areas, parking changes, marina relocation, dredging, sediment, or underwater construction constraints.
After checking sources
Check and recalibrate
The major correction is ramp slope. A 30-degree ramp is extremely steep for launching boats. Boat-ramp guidance commonly uses about a 12% to 15% slope, meaning 12 to 15 feet of vertical drop per 100 feet of horizontal run. That is only about 7 to 8.5 degrees.
vertical drop ~= 33 ft
~= 1 x 10^1 m
boat-ramp slope ~= 12% to 15%
~= 0.12 to 0.15
horizontal run ~= vertical drop / slope
~= 33 ft / (0.12 to 0.15)
~= 2.2 x 10^2 to 2.8 x 10^2 ftAt these shallow angles, the sloped surface length is almost the same as the horizontal run. So a good ramp-length estimate is about 225 to 280 feet, or roughly 70 to 85 meters. Use 250 feet, or about 75 meters, as the memory-scale answer.
Now estimate concrete. A two-lane ramp might be roughly 25 to 30 feet wide, or about 8 to 9 meters. Cast-in-place concrete ramps are often around 6 to 8 inches thick, or about 0.15 to 0.20 meters, before separately thinking about subbase and reinforcement.
middle case:
ramp length ~= 75 m
ramp width ~= 8 m
slab thickness ~= 0.18 m
concrete volume ~= (75 m) x (8 m) x (0.18 m)
~= 1.1 x 10^2 m3A reasonable Fermi range is about 70 to 170 m3 of concrete, depending on width and thickness. Since 1 m3 is about 1.3 cubic yards, that is about 90 to 220 cubic yards.
Matt's density estimate was high. Normal-weight concrete is more like 2.3 to 2.4 x 10^3 kg/m3, about a quarter of iron's density.
concrete mass ~= volume x density
~= (70 to 170 m3) x (2.4 x 10^3 kg/m3)
~= 1.7 x 10^5 to 4.1 x 10^5 kgThat is about 170 to 410 metric tons of concrete, with a middle estimate near 260 metric tons. The lower volume and lower density errors partly cancel, which is why Matt's 600-ton estimate is high but not absurdly far away in order-of-magnitude terms.
Ready-mix trucks are often around 9 to 11 cubic yards. That puts the ramp extension at roughly:
truckloads ~= (90 to 220 yd3) / (10 yd3/truck)
~= 9 to 22 truckloadsFor cost, separate concrete material from installed civil work. At roughly $160 to $195 per cubic yard, the ready-mix concrete alone is about:
concrete material cost ~= (90 to 220 yd3) x ($160 to $195/yd3)
~= $1.4 x 10^4 to $4.3 x 10^4The installed cost could easily be higher after grading, subbase, rebar, forms, drainage, shoreline work, mobilization, and site constraints. A simple slab square-foot estimate puts the work in the tens of thousands of dollars, and a remote reservoir ramp extension could plausibly become a six-figure job.
A 250-foot ramp extension is about 80% of a 100-yard football field, not counting end zones. So the answer is not a megaproject, but it is also not ordinary driveway paving. For one ramp, it is a substantial small civil-engineering project; multiplied across marinas and repeated water-level drops, it becomes an ongoing infrastructure adaptation problem.
Post-check reflection
Matt's reflection
I lucked into an okay answer because of canceling errors, which happens often in Fermi problems. I was way off with the angle: my mental model used a slope about double what I should have used, which made the required ramp much shorter than it really is.
I also complicated things with the triangle system because I was trying to complete the problem using mental math, no calculator, so I could not just use a trig function like sine. I was way off on the concrete density estimate too. That is worth keeping track of moving forward: concrete is about a quarter the density of iron.
Finally, I assumed a half-meter-thick slab, but it should have been more like one-tenth to one-quarter of a meter. I also missed the ask about concrete-truck loads. It was an off day. As far as the news item itself goes, all the numbers seem about right.
Recommended memory peg
Remember boat ramps are roughly 12% to 15% grade, concrete is about 2.4 x 10^3 kg/m3, and a ready-mix truck carries about 10 cubic yards. For shallow ramps, use added length ~= vertical drop / slope.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.