Steamboat Hill

Copyright 2009 by Paul Niquette.  All rights reserved.

ith a current of four knots, any steamboat heading down river could in principle 'make good' that much speed without burning even one lump of coal.  Hooray for economy and ecology.  Oh, but the vessel must be kept away from shores and shoals.  Assuring rudder authority required a minimal amount of steady headway. One knot ought to be enough, making good five knots down river between ports-o'-call.

Heading up river is a different matter.  Our puzzle calls for the vessel to maintain some unspecified speed against the current.  Some solvers may be inclined to note that a speed slightly faster than four knots would assure the least expenditure of energy. Indeed that minimizes coal consumption per unit of time; however, the vessel would hardly be moving, which effectively maximizes coal consumption per unit of distance-made-good.  Commanding a higher speed would increase coal consumption per unit of time, while shortening the time enroute, which might seem to decrease coal consumption per unit of distance-made-good.  Or would it?

Paraphrasing an observation made in A Certain Bicyclist...
 

For all things that move through fluids -- bikes and birds, buses and bullets, barracuda and boats -- there's always the invisible enemy, drag.

Drag typically increases with the square of speed.  Power, steam or otherwise, required to overcome drag also increases with speed, thus power increases with the cube of speed.  Have a look at this table...
 

 
Speed 
(knots)
 Making 
 Good 
Drag
 (relative) 
 Power
 (relative) 
Power
(% maximum)
 
1
-3
1
1
0.1
 
2
-2
4
8
0.6
 
3
-1
9
27
2.0
 
4
0
16
64
4.8
[a]
5
+1
25
125
9.4
[b]
6
+2
36
216
16.2
[c]
7
+3
49
343
25.8
 [d] 
8
+4
64
512
38.5
 
9
+5
81
729
54.8
 
10
+6
100
1,000
75.1
 
11
+7
121
1,331
100.0

The four alternatives in the puzzle offer a range of 1.6:1 in 'hull' speeds (8/5), which imposes a 4:1 ratio (512/125) in requisite power and ironically 'makes good' a 4:1 speed ratio (+4/+1). Sophisticated solvers will probably proceed as follows:

Let p = k v3
...where p = required power, k = constant of proportionality, v = hull speed (knots).

or the Steamboat Hill puzzle, the units for p in the 21st Century might be 'horsepower' as commonly used in agricultural, automotive, and aviation parlance; however, more likely in the 19th Century, p was measured as the rate of coal consumption in pounds per hour.  What we seek to maximize is the distance-made-good per pound of coal consumed (analogous to 'miles-per-gallon') by writing...

x = (v - c) / p
...where x = nautical miles-made-good per pound of coal consumed and c = speed of the opposing river current in nautical miles per hour.

Thus, x = (v - c) / k v3, the maximum of which can be found by elementary Differential Calculus: setting the first derivative dx / dv to zero and solving for v...

dx / dv = [k v3- 3 k v2 (v - c)] / [ k2 v6] = 0
...giving us the value of v for any value of c...
v = 3 c / 2
...so that for c = four knots. the most fuel efficient value for v is ...
 
 [b]   v = six knots

...which can be confirmed by taking ratios of entries in the table above...

[a] 1 /  9.4 =  0.106 
[b] 2 / 16.2 = 0.123 <== Most fuel efficient. green spin
[c] 3 / 25.8 = 0.116
[d] 4 / 38.5 = 0.104
Solvers may note in passing these two ironic properties of the Steamboat Hill solution...
  1. That for any vessel, neither the maximum available power nor highest attainable performance can participate in the analyses of the conservation case. 
  2. That the faster the opposing river current the faster the most efficient hull speed and the faster the speed-made-good (for c = 6 kts, say, v = 9 kts, and v - c = 3 kts).
If you find others, please send them to puzzles@niquette.com.
 
 


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