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1) So how much wood did it take to build that canoe? (Assume "juice can" shape - flat bow and stern - and a wood weight of 40 pounds / cubic foot. Choose a board thickness sufficient to withstand crocodile jaws, and note that thickness in your written calculations.) To check your results (and the wood-weight estimate): Prof. Fischer's favorite canoe is 13 feet long and pointed at both ends, with the curve starting almost at midlength. Its hull is 1/4" thick, but there is some extra weight on the shape (thicker wood, some brass). But he can carry it balanced on one shoulder (shoulder at midlength of gunwales). Second check: Can you carry, balanced on one shoulder, a ten-foot long 4"x4" wooden post of the kind used to build backyard garden structures? The volume of the wood on the big Humboldt canoe can be calculated by calculating the volume of two cylinders, one inside the other. The outer cylinder is the outside surface of the hull. The inner cylinder is the inside surface of the hull. By class agreement, 4 inches of wood may deter the crocodiles for a while. The inside cylinder has a width of 3 feet (or 36 inches). The outside cylinder has a width of 3'8" (or 3 2/3 feet or 44 inches or approximately 3.67). The canoe is 30 feet long. Be sure to do your calculations entirely in feet (and fractions of a foot) or entirely in inches, though eventually you want to get the wood volume in cubic feet. Here we'll stay with feet. Humboldt's canoe had a semi-circular cross-section. So picture it as a cylinder 15 feet long which is then split down the middle (lengthwise). Put the pieces end to end and you get a 30-foot canoe with a semi-circular cross section. Volume of cylinder = area of its base multiplied by its height The base is a circle. Its area is calculated by the formula: (radius squared) x pi. Pi is approximamtely 3.14. The radius of a circle is 1/2 of its diameter. So the radius of the inner circle of the canoe's stern/bow is 1.5. The square of the radius is 1.5 x 1.5 = 2.25 So the area of the base of the inner cylinder = 2.25 x 3.14 = 7.065 (round that to 7) The volume of the inner cylinder is 7 x 15 = 105 cubic feet. The radius of the outer circle is 1/2 of 3.67 = 1.84. The square of 1.84 = 3.3856. Call it 3.4 (which will make sure there will be enough wood). So the area of the base of the inner cylinder = 3.4 x 3.14 = 10.2 (rounding up to make sure there's enough wood). The volume of the outer cylinder = 10.2 x 15 = 150 cubic feet. The volume of the hull is 150-105 = 45 cubic feet. If a cubic foot of Humboldt's wood weight 40 pounds, then the hull of the canoe (without any extra fittings, or mast & sail, or whatever) is 40 x 45 = 1,800 pounds. Ohhhh!, We forgot the weight of the bow and stern pieces. The two pieces (semi-circles) could be made from a single circle of wood that is 3 feet in diameter and 4 inches thick. area of that circle = 7.065. Its thickness is 1/3 of a foot. So the volume of the stern/bow circle is about 2.35 cubic feet. It would weight (2.35 x 40) about 94 pounds. So the entire hull (just the hull, no extra fittings) would weight close to 1900 pounds (slightly less than an American ton). If it had a crew of 12 people who wanted to pick it up and carry it, each would have to lift close to 160 pounds - which would probably be about what those crew members would each weight after a very full meal. Let's figure the volume of Prof. Fischer canoe as though it were built in the same shape as Humboldt's (semi-circular hull, with flat bow and stern transoms): The beam (breadth/width) of his canoe is at most 2 feet. So the volume of the inner cylinder is calculated by these steps: radius = 1; square of radius = 1; area of base of cylinder = square of radius x pi = 3.14 square feet (for the bow and stern transoms combined); volume of inner cylinder = 3.14 x 6.5 feet = 20.41 cubic feet (less than 1/5 the volume of Humboldt's canoe). The volume of the outer cylinder is calculated by these steps: radius = 1 foot + 1/4". One-fourth of an inch = 1/48 of a foot or .02 of a foot. radius = 1.02; square of radius = 1.02 x 1.02 = 1.0404; area of base of outer cylinder = 1.0404 x 3.14 = 3.267; volume of outer cylinder = 3.267 x 6.5 = 21.2346 cubic feet. Volume of the wood in that little skinny-hulled canoe (no protection at all against crocs) = volume of outer cylinder = volume of inner cylinder = 21.2346 - 20.41 = 0.8246 cubic feet The weight of that hull = 40 pounds x 0.8256 = 32.984 pounds. Add the blunt semi-circular bow/stern transoms by making a circle 2 feet in diamer and 1/4 inch thick: area of that circle is 3.14 square feet; its thickness if .02 feet; the volume of that disk = 0.0628 cubic feet; the disk weighs 0.0628 x 40 pounds = 2.5 pounds. All together that little skinny croc-endangered canoe (wooden hull without additional fittings) weighs a little more than 35 pounds (32.984 + 2.5). To check that figure: The actual canoe gradually narrows to almost nothing at both ends (but retains its same height as at the middle). With no horrible strain Prof. Fischer can pick it up and hook the mid-point of its gunwales (side) over one shoulder and then walk a considerable distance with the canoe balanced on that should, one end quite a ways in front, one end quite a ways in back (he watches carefully for people and cars, especially when he turns), and the canoe hull sideways (open side - top, when in the water) toward himself, with the other gunwale pressing against him mid-thigh. The canoe does have some more wood on it - small decks fore and aft, two thwarts (cross-braces), and some metal (keel strip, screws). He usually carries one double-ended kayak paddle (est. weight 3 pounds) and one backup paddle (1.5 lbs.). Call it 40 pounds. A gallon of milk weight 8 pounds (one gallon = 4 quarts = 8 pints, and "a pint's a pound, the world around"). Prof. Fischer can fairly easily carry two gallons of milk in each hand, hanging down from his hands and with most of the weight therefore on his shoulders. So, yes, 40 pounds is a likely figure for the weight of that little skinny canoe before the crocs chomp it up and swallow it. 2) Using the activity we did in class several weeks ago, recalculate the volume (inner) of Humboldt's canoe, with the same inner width (3 feet), but with a conical bow taking up the front 10 feet of the 30-foot entire length. Slight change in problem: Make both the bow and the stern pointed, so that it can be maneuvered in either direction with equal ease in those fast crocodile-infested currents. So the canoe will consist of 10 feet in the middle shaped like a semi-cylinder, and 10 feet of (semi-)cone-like shape at both bow and stern. Imagine a cylinder 5 feet high with a conical "hat" on it that is another 10 feet high. To get the canoe shape we'll split that shape down the middle vertically. The volume of the cylinder (see calculations already done above) = area of its circular cross-section (7.065 square feet) x 5 feet = 35 cubic feet Now remember (or, as Prof. Fischer had to do, look up) the formular for the volume of a cone: http://math.about.com/od/formulas/ss/surfaceareavol_2.htm (Prof. Fischer thought he remembered it as 1/3 the area of the base multiplied by the height, for both a pyramid and a cone, and he was right!). So the area of the base of the cone is So the volume of the cone = (7 x 10)/3 = 23 cubic feet. The volume of the stubby cylinder with its cone "hat" = 35 + 23 = 58 cubic feet. And that would be the volume of the canoe with pointed bow and stern. That's only a little more than half the volume of the flat-ended cylindrical canoe. The pointed canoe would be much more maneuverable, but it would almost as much wood for its construction and would therefore weigh almost as much as the clunky flat-ended canoe. It probably would be less stable (more likely to capsize), at least on flat water with no wind. Building things like that is always a matter of trade-offs. 3) Humboldt's map of the Casiquiare (article and large-scale drawing), which (hint!) is very near the equator, was 3 degrees of longitude in error. How big was that error, in terms of miles and percent inaccuracy and in terms of what may have caused it: a chronometer or other calculation of time that was used to determine longitude? Extra glory (not extra credit): How serious an error, in terms of Humboldt's claim to scientific precision in his own time, would have been an error of 3 degrees of LATITUDE? The diamer of the Earth is approx 8000 miles. It's circumference is therefore about 25,000 miles. Round that to 24,000 to make calculations easier. The Earth rotates once a day. A day is 24 hours. Therefore a point on the Earth's equator rotates 1000 miles per hour as the Earth turns. A circle has 360 degrees. That point on the Earth's equator therefore rotates 15 degrees per hour. In other terms, one degree of rotation happens very 4 minutes (60 minutes / 15 degrees). If Humboldt were using a chronometer (clock) as part of calculating longitude, a clock error of 12 minutes (from rolling seas, rusty gears, a fly in the mechanism, whatever) would produce a longitude error of 3 degrees. That's not surprising. The error might have come, otherwise, from small problems in the astronomical tables Humboldt might have used as an alternate to determining local time and then comparing it his original time (probably the longitude of Paris). In terms of mile: a degree of longitude at the equator = nearly 70 miles (25,000 miles / 360 degrees). Three degrees of error = 210 miles. If Humboldt later discovered the error (as he probably did), he would still have been able to correct his maps, as long as they had no internal errors. A mistake of 3 degrees of LATITUDE would have been an indication of serious flaws in observation or calculation, since at the time latitude was determined by astromical events that could be observed relatively easily (at least in good weather and on land that has a flat horizon): an angular measurement of the height of the North Star above the horizon (though it gets hard to see the North Star when you get near the equator, and it can't be seen south of the equator, where there is no clear Pole Star); or a similar measurement of the height of the sun about the southern horizon at noon. |