# Part 5: What Moves You

by Shawn Burke, Ph.D.

When you paddle, each stroke propels your hull forward against our familiar nemesis, drag.  But why does the canoe move forward, and not in some other direction?  And more fundamentally, why does it move at all?  The concept of momentum provides the answer.  And it all can be understood by considering space boots.

When I was in 6th grade I bought a book of science fiction stories at a garage sale.  One of the tales centered on an astronaut marooned outside his spacecraft; how he became marooned, I don’t recall.  Fortunately, he was wearing a space suit – it’s pretty cold in outer space.  Unfortunately, the space suit had no thrusters to propell him back to his craft.  And he had a limited supply of air left to breath.  How did he get back inside his rocket ship before running out of air?  By using science!

Our stranded astronaut recalled that when you fire a rifle, there is recoil.  The rifle flings a bullet in one direction, and in response the rifle moves in the opposite direction.  The astronaut realized that if he threw something in the direction opposite his spacecraft, his body would recoil in response and move toward the ship.  Now he couldn’t throw his oxygen tanks (gotta breathe), his space helmet (thus avoiding the worst case of brain freeze ever), or his space gloves (he needed functioning hands to open the hatch back into his spacecraft).  So he took off his space boots and threw them in a direction away from his rocket ship as hard as he could.  He had a pretty good aim because he drifted slowly toward the ship’s hatch, and was able to climb back inside.  All because he remembered physics, and the concept of conservation of momentum.  And, of course, because he had space boots.

We’ve all heard the word “momentum,” but what does it means in scientific terms, and how does it apply to paddling? Momentum can be thought of as “mass in motion.” A hull and its paddler(s) have mass, so if the hull is moving then it has momentum. An object’s momentum depends upon its mass and its velocity. Formally, momentum – typically written in shorthand as “P” – is equal to the mass of the object times its velocity:

$Momentum\equiv P=mass\times velocity$

This equation shows that momentum is linearly proportional to an object’s mass, as well as linearly proportional to its velocity.  And by linearly we mean that neither quantity appears as a square, cube, square root, etc., but only as itself.  This means that if you throw a baseball at 45mph (72.4 kph), and a major league pitcher throws the same baseball at 90mph (144.8 kph, e.g., twice as fast), then the baseball thrown by the major leaguer has twice as much momentum as when you threw it.

So now consider a hull paddled by a solo paddler.  We’ll adopt a hull-centric coordinate system. At the end of each stroke their hull has a certain amount of added momentum Phull, which is quantified as the total mass of the hull mhull – including the mass of paddler – times the hull’s change in velocity vhull.  Note that we’re concerned with changes in momentum with each stroke; a given stroke doesn’t contribute to the hull’s accumulated momentum from all previous strokes.  Consequently,

${{P}_{hull}}={{m}_{hull}}\times {{v}_{hull}}$

With each stroke the paddler imparts an increment of momentum to the hull above its current steady-state.  This is because the paddle blade accelerates and “throws” a volume of water.  In an inertial, Earth-reference coordinate system your paddle appears to be planted in concrete at the catch, albeit with a little slippage.  Yet if you watch the strokes of adjacent paddlers while underway you’ll note little pools of swirling, aft-moving water behind their blades at the conclusion of their power phase.  Each stroke changes the momentum of the water entrained by the paddle.

Figs. 1-5 show a side view of the power phase via a sequence of five photographs.  The paddle rotates during this phase.  It stays fairly well planted while it is near the vertical, slipping slightly sternward in the inertial (Earth-based) reference frame.  More importantly, in the hull-centric frame of reference the blade moves a significant distance along the hull – note the number of white fiducial lines the paddle traverses during the stroke compared to the number in the inertial reference frame atop the water.  From the paddler’s point of view the blade has displaced a mass of water sternwards.

Figs. 1-5: Power phase, side view with hull and fixed fiducials.

Your paddle stroke imparts momentum Pwater to an entrained mass of water mwater, which we idealize as a plug of water having a velocity vwater, in the direction opposite your hull’s motion[1]. The momentum of this idealized plug of water is then

${{P}_{water}}=-{{m}_{water}}\times {{v}_{water}}$

The minus sign indicates that the water’s momentum is in the opposite direction of the hull’s momentum.  This is because in physics velocity is a vector quantity: velocity has both a magnitude (referred to commonly as speed) and a direction.  The minus sign here indicates that the displaced water and the hull move in opposite directions.

Physics teaches us that the momentum of a system is conserved. Here, the system is the hull and paddler, as well as the slug of water “thrown” by the paddle. Consequently, at the conclusion of each stroke the total momentum of the system – here, the hull/paddler(s) and the paddled water – must equal zero. This may be written as,

${{P}_{hull}}+{{P}_{water}}={{m}_{hull}}\times {{v}_{hull}}-{{m}_{water}}\times {{v}_{water}}=0$

As a result, the momentum of the hull plus paddler, plus the momentum of the slug of water, must be equal. This means

$\Rightarrow {{m}_{hull}}\times {{v}_{hull}}={{m}_{water}}\times {{v}_{water}}$

So to move the hull, you impart momentum to the water entrained by your paddle. And the hull will travel in the direction opposite the water that your paddle has “thrown.” This should be familiar to anyone who has paddled a canoe, kayak, or SUP; the hull doesn’t move in the same direction as your paddle blade moves in the water, nor does it move perpendicular to the path of the paddle, because momentum has both a magnitude and a direction and the system’s momentum must be conserved.

Perhaps the most interesting take away, particularly relevant to our discussion, is that the momentum of the paddled water equals the momentum of you and your hull at the end of each stroke. It’s exactly like our astronaut floating freely in space – to move toward his spacecraft, the astronaut threw his space boots in the direction opposite his ship, and off he zoomed. If he had heavier space boots, or threw them faster, he could move faster in response.

So what does this tell us about moving faster? The equation above can be expressed in terms of hull speed vhull as

${{v}_{hull}}=\frac{{{m}_{water}}\times {{v}_{water}}}{{{m}_{hull}}}$

To increase hull speed, we see from this equation that there are several options:

• For a fixed paddle size and power phase stroke finishing speed, the numerator in the equation is a constant. So, to go faster our paddler should get a lighter hull, lose weight, or both since the only remaining variable in this case is the combined mass of the hull and paddler.
• For a fixed power phase blade speed, our paddler could move a bigger slug of water by using a paddle with a bigger blade in order to entrain more water with the stroke.
• For a fixed paddle size, our paddler could move the entrained water faster with a faster  power phase blade speed, or use a smaller paddle but have an even faster power phase speed.
• Or, our paddler could lose weight, buy a lighter hull, get a paddle with a bigger blade, and move that blade faster. Phew! I’m tired already, not to mention broke from the cost of all the new gear.

Practically speaking, each option will appeal to specific paddlers based upon energy, paddle force, and power considerations. We’ll address energy issues below; paddle force and power are addressed in Part 11: About the Bend, Part 1: Tandem vs. Solo, Part 9: Power to the Paddlers, and Part 25: Impulse, Part 1.

Now what if the canoeist isn’t paddling, but instead is poling? As you may (or may not) know, canoe polers “stand tall in their canoes” and propel themselves using a pole of wood or aluminum, driving the pole into the riverbed and pushing off as illustrated by 12-time ACA poling National Champion Harry Rock in Fig. 6.

Fig. 6: Poling up a drop.

Assuming the pole is well planted and doesn’t slip, the poler is pushing against the mass of the Earth – a pretty large mass. In this case our equation for momentum conservation takes a particularly interesting form:

${{P}_{hull}}+{{P}_{Earth}}={{m}_{hull}}\times {{v}_{hull}}-{{m}_{Earth}}{{v}_{Earth}}=0$

So, each time the poler plants their pole and propels their canoe forward, they are changing the momentum of the Earth, pushing it in the opposite direction!

As noted above, a hull can attain a particular speed by using the paddle to move a larger mass of water slowly, or a smaller mass of water more quickly. The choice is yours as long as total momentum is conserved. The difference between these two scenarios becomes more apparent when we consider the energy expended in each scenario, e.g., the cost in energy to you, the paddler.

At the end of each stroke, the energy in the hull/paddler/water system is due in the motion of the hull and paddler(s) and the motion of the paddled water. This motion energy is called kinetic energy. Kinetic energy is linearly proportional to mass, and proportional to the square of velocity[2]. The total kinetic energy of the system is then expressed as the sum of the kinetic energy of the hull and paddler, and the kinetic energy of the water:

${{E}_{kinetic}}=\frac{1}{2}{{m}_{hull}}v_{hull}^{2}+\frac{1}{2}{{m}_{water}}v_{water}^{2}$

Let’s consider the impact of paddle blade size for a given hull speed vhull. We’ll consider two cases: a smaller paddle blade with an area represented by A1, and a bigger paddle blade with an area represented by A2. We assume that with each stroke the paddle pushes a plug of water having a volume covering the area of the blades, extending a distance T behind the power face – this is merely a simplifying assumption for the purposes of comparing the impact of blade size, as the detailed hydrodynamics of blades is a good deal more complex. This means that at the end of the stroke the paddler moves a mass of water equal to the volume of water (the blade area A time the thickness T) time the density of water $\rho$:

${{m}_{water}}=A\times T\times \rho$

For the smaller bladed paddle, the total system kinetic energy at the end of the stroke is then

${{E}_{1}}=\frac{1}{2}{{m}_{hull}}v_{hull}^{2}+\frac{1}{2}\rho {{A}_{1}}Tv_{water,1}^{2}$

For our bigger bladed paddle, the total system kinetic energy at the end of the stroke is

${{E}_{2}}=\frac{1}{2}{{m}_{hull}}v_{hull}^{2}+\frac{1}{2}\rho {{A}_{2}}Tv_{water,2}^{2}$

The ratio of these two energies, which lets us determine how much energy is expended by using a smaller paddle vs. using a larger paddle to attain the same hull speed, is

$\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{\frac{1}{2}{{m}_{hull}}v_{hull}^{2}+\frac{1}{2}\rho {{A}_{1}}Tv_{water,1}^{2}}{\frac{1}{2}{{m}_{hull}}v_{hull}^{2}+\frac{1}{2}\rho {{A}_{2}}Tv_{water,2}^{2}}$

For the sake of comparison, we’ll assume that the kinetic energy of the hull is the same for both blade sizes.  This means that the first terms in both the numerator and denominator of the equation above are the same; the difference lies only in the respective kinetic energies of the water moved by the small and large blades.  The ratio of these energies is

$\frac{{{E}_{water, 1}}}{{{E}_{water,2}}} = \frac{{{A}_{1}}v_{water,1}^{2}}{{{A}_{2}}v_{water,2}^{2}}$

One next uses the momentum conservation equation for each scenario to solve for vwater,1 and vwater,2 in terms of the blade areas. After a little bit of algebra, the approximation for the ratio of energies above simplifies to become

$\frac{{{E}_{water,1}}}{{{E}_{water,2}}} = \frac{{{A}_{2}}}{{{A}_{1}}}$

Since the larger blade’s area A2 is always bigger than the smaller blade’s area A1, the ratio A2 / A1 will always be greater than 1. The approximation above shows that you will always expend more energy paddling with a smaller blade than paddling with a larger blade in order to attain the same hull speed. The cost of using a smaller paddle to achieve the same hull speed is greater energy expenditure – by you. Surprise!

In both scenarios the mass and speed of the canoe and paddler at the end of the stroke stays the same, hence their contribution to kinetic energy is fixed. But while the mass of water moved by the smaller paddles is less, the water’s kinetic energy is only linearly proportional to this reduction.  This gain is more than offset by the squared dependence of kinetic energy on the velocity. Since the smaller mass of water moved by the smaller blades must be moved faster to achieve a given hull speed because of momentum conservation, what you appear to gain in moving a smaller mass of water you more than lose in energy expenditure by having to move it with a faster power phase speed. So if you prefer a smaller blade, use it briskly, and have a well-trained aerobic/anaerobic engine.