What is the fastest time possible?

As far as I know, the fastest regatta boat timed over a 10 foot raingutter was raced by Robert Defraitas in Ovilla, Texas, in October, 1999. His opponent, only inches behind, was his wife, Cathy. Independently clocked by myself and the regatta coordinator, both stopwatches read 1.86 seconds!

Robert's 10 year old daughter, Amy, won the 3-6th grade division in 2.44 seconds. Again, attested to by two clocks and two separate races!

Is this as fast as we can expect a raingutter regatta boat to go? First of all, both boats were hydroplanes which means they were able to skim across the surface of the water with little friction. Second, they were one ounce boats. Our rules had previously required boats to weigh at least 2 ounces. Only a few older children could make them hydroplane for a couple feet at a time. Lowering the weight minimum allowed most children who built hydroplanes to sustain hydroplaning nearly the whole race.

Measuring Up

To find out some of the dynamics of these fast boats, we can use the hydroplane model. Though the model derived a relationship between the breath speed and the coefficient of sliding friction, we will find it difficult to use the race to measure either! Never-the-less, we can find out the speed anywhere along the boat's path from just making two simple measurements during the race.

First, we need to measure some parts of the boat and find some good values to use for coefficients and physical properties. The table in the profile tables indicate some reasonable values for the variables used in the hydroplane models.

Along with these model variables, we need at least two measurements; time, t, and distance, L, travelled during that time.

Terminal Velocity

Let's use the best time I know of, t = 1.86 seconds for a 10 foot course, L = 114, the standard path length. Assume that Robert blew his boat across in one breath (running along side to stay close). The average speed of the boat is then

Average speed = L/t = 114/1.86 = 61.3 in/s

The terminal velocity must be greater than this average speed.

vt > 61.3 in/s

To find out what the terminal velocity was for that boat in that race use the model equation for terminal velocity that doesn't need the breath speed which we don't yet know.

vt = m arccosh(exp(2epA(y=L)/m))/2epAt

First, evaluate epA once so we can use it twice,

epA = 0.7*0.000001878*7.5 = 0.00000986 ozs2/in2

Now cascade the results so it can be checked more easily to be sure it's done right.

vt = 0.00259 arccosh(exp(2*0.00000986*114/0.00259))/(2*0.00000986*1.86)

vt = 0.00259 arccosh(exp(0.8679))/0.000036677

vt = 70.6158 arccosh(2.3820); arccosh(X) = ln(X+\[XX-1])

vt = 70.6158 ln(2.3820+\[2.3820*2.3820-1])

vt = 70.6158 ln(2.3820+2.1619)

vt = 70.6158*1.5138

vt = 106.90 in/s almost 9 feet per second!

Finish Line Speed

Now that we know the terminal speed, we can find the speed at any point in the race assuming Robert was able to maintain the same breath force the whole way! In particular, let's find out what the final speed was. From the model we have,

v(y) = vt \[1-exp(-4epA(y=L)/m)]

Cascading the evaluation, we get,

v(114) = 106.9 \[1-exp(-4*0.00000986*114/0.00259)]

v(114) = 106.9 \[1-exp(-1.7360)]

v(114) = 106.9 \[1-0.1762]

v(114) = 106.9*0.9076

v(114) = 97.02 in/s just a tad over 8 feet per second.

So the final speed was about 91% of the terminal speed.

How Much Breath?

How much breath was needed to set this reigning raingutter regatta record? To give us a clue, we'll evaluate the model relation between breath speed and sliding friction. Results from the simple equation can be put into a table to give us a feel for the trade-off involved. However, we know from measurements that Robert's breath speed was about 500 inches per second! So we will be able to estimate the friction coefficient for sliding on water.

Start by evaluating some repetitive factors,

epA = 0.7*0.000001878*7.5 = 0.00000986 ozs2/in2

spS = 1.3*0.000001878*5 = 0.00001221 ozs2/in2

B(n) = \[(epAvt2+2nmg)/spS]

Cascade the evaluation to avoid errors,

B(n) = \[(0.00000986(106.9)2+2n)/0.00001221]

B(n) = \[(0.1127+2n)/0.00001221]

B(n) = \[9230 + 163800n]

This simple equation can be evaluated for different values of the sliding friction coefficient to see how fast the breath must be to make the boat go the distance in 1.86 seconds.

Note that we can solve for n if we know B,

n = (B2-9230)/163800

Assuming B = 500 in/s,

n = (250000-9230)/163800 = 1.5. mult by 1 oz = breath force means no motion!

n = 0 would closely model the boat scooting across ice. It gives a lower bound on the breath force spSB2/2. It works out to be 0.056 oz which seems too small. Blowing with a speed of 205 in/s (n = 0.2) gives a force of 0.26 oz, about 5 times stronger at only twice the speed.

Amy's Race

Amy, loaned dad her boat for his race. If we use Amy's time of 2.44 seconds over 114 inches, we can find the same information about Amy's race. But we won't have to compute everything because we cascaded the computations above. Without much work we can fill in the following tables,

Notice that Amy's boat speeds are 76 percent of her dad's. Let's see how their breath speeds compare.

As can be expected from the expressions for their breath speeds, Amy's percentage of her dad's breath speed increases as the sliding friction coefficient increases. Presumably, since the same boat was used in both races, the sliding friction coefficient, n, would be the same. But what is its value?

We know Amy's blew her boat to 76% the speed that her dad did. That's a kinetic energy of m(0.76v)2/2, where v is her dad's final race speed. Factoring out the percentage gives Amy's boat's kinetic energy in terms of how much her dad's had. That's 0.58mv2/2, so she gave the boat only 58% as much energy as her dad did.

That energy was transfered from her breath's kinetic energy to the boat. So her breath was transfering only 58% of the energy her dad's was throughout the race. Amy's breath had to have energy ma(0.76B)2/2 where ma is the mass of air and B is dad's breath speed. So all things being equal, Ami had to blow her breath about 76% as fast as her dad!

Given that Amy's breath speed is 76% of her dad's, we can see from the breath speed comparison table that the difference is 76% only at the zero value for the coefficient of sliding friction, n = 0. There are a few factors that suggest Amy's preformance could be more than 76%. If this is correct, this result is truly remarkable! It says that when a small boat hydroplanes, it acts as if it is on a frictionless surface!

But suppose Amy took 1 breath more than her dad, the "down" time would have to be accounted for and the estimated speed of her breath would have to be increased to get the same time over the course. That would mean a precentage greater than 76% and a non-zero coefficient of sliding friction. The only way to know the coefficient and its nature for sure is to make some breath speed measurements. But its good that we don't need to know what it is to find the boat speeds!

Note the breath speeds for both Amy and her dad are just a bit lower than their finish line speeds. The fact that a regatta boat can sail faster than the wind behind it shouldn't surprise you. It's like a rocket that goes faster than its thrust.

So what is the fastest a regatta boat can go?

About 10% faster than you can you blow! (if everything goes right in one blow)

But the length of the raingutter keeps it to just over how fast you can blow.