Physics of Stream/Depth & Rowing
Recent Modifications
 20JAN08 Updated layout.
1. Introduction
The creation of this page was prompted by discussions on the
rec.sport.rowing newsgroup about why
rowing upstream or downstream feels different, and also the difference
between deep/shallow water.
I'll start by saying I'm assuming that rowing downstream feels 'heavier' than
rowing upstream, which is my personal impression although some would argue
it's the other way around  which I can't explain except as a purely
psychological effect.
Also, I'm discounting the change in air resistance,
i.e. when you row downstream you're always moving faster relative to the air
than rowing upstream, so relatively speaking, downstream pieces are into more
of a headwind than upstream pieces. This is clearly correct but, I think,
negligible  again speaking personally, I reckon that I can tell the difference
between increased water resistance and increased air resistance, and the
upstream/downstream difference feels like water resistance to me.
2. Viscosity
Consider two surfaces separated by fluid, a distance H apart,
with the upper surface moving at V and the lower surface fixed.
The fluid adjacent to the upper surface will be dragged along and also be
moving at velocity V while the fluid adjacent to the lower surface
will be stationary, so that a constant velocity gradient V/H
(also known as 'shear') is set up in the fluid.

Figure (2.1)

As any physics textbook will tell you, the Resistance R
(measured as force per unit area) caused by viscosity is given by:
where e is the coefficient of viscosity (measured in kg/m/s,
assumed constant), and the shear dv/dz = V/H
in this case, so
This tells you that viscous drag (resistance) on the upper surface increases in
proportion to velocity. However, this is only really applies to situations
where the horizontal lengths are much larger than the separation H,
so that the shear layer is constant along the whole length.
This is only true for boats rowing over
very shallow (inches) water. For most purposes, a better model is required.
3. Boat Resistance
As a boat moves through stationary water, the water in contact with the bows is
immediately accelerated to the boat speed V, but the shear layer
can only grow downwards at a fixed speed W (set by the mean free
path of molecules). So the lower (static) boundary of the shear layer slopes
downwards from the bows:

Figure (3.1)

Below point x along the hull, the boundary layer will have been
growing for a time t=x/V, so will have reached a depth
h=W.t=W.x/V. So using Eq.(2.1),
the viscous drag at point x is given by:
(3.1)
 R(x) = e.V^{2}/(W.x)

This is the origin of the V^{2}
law for boat resistance (cf.
resistance proportional to V in previous section)
 see Section 2 of Basics
4. River Flow
River flow is driven by the hydrostatic pressure gradient, which is constant
across the whole
cross section of the river. Were it not for viscosity effects, this would mean
that the stream flowed at an equal speed at all points within the crosssection
since each point is driven with the same force. However, due to viscosity, the
flow is slower near the fixed boundary (riverbed and banks) and faster near
the free boundary (surface, since the air offers relatively little resistance
to flow), and the quickest flow will be the furthest from the fixed boundary,
which means away from the sides and where the river is deepest.

Figure (4.1)

The diagram shows a crosssection of flow over an uneven river bed, the
contours represent flow speed.
Flow is 0 next to fixed boundary, 1
for 1 layer away, and so on. As with most rivers, being wider than deep, the
flow rate in most places is determined by the depth rather than the distance
from the sides. The surface flow is therefore quickest (3) over the
right hand channel and slowest tucked right in to the sides, or over the
central ridge.
5. Effect of River Flow on Resistance
If there is any flow, the stream will have its own vertical shear
(left diagram in Fig. 5.1). From
Section 3, a boat moving through still water will also
set up its own shear layer at some fixed point beneath the hull
(right diagram).

Figure (5.1)

When we have a boat moving in a stream, these two shears will sum
as in Fig. (5.2)

Figure (5.2)

The left diagram shows the shear set up by a boat moving upstream
at speed V relative to the water, which is itself moving at speed U relative
to the bank or the riverbed. In this case there is some cancellation
between the two shear layers: the velocity shear under the boat is
reduced so there is less apparent drag compared to the still water case.
The right diagram shows the opposite situation for a boat moving downstream.
Here the shear and apparent drag are increased.
In general, the faster the flow, or the shallower the water, the greater the
shear so the greater the difference in resistance.
6. Shallow Water Resistance
The difference between sections 2 and 3 was that in the former case the shear layer had fixed
depth whereas in the latter it grew continuously as the boat passed over.

Figure (6.1)

In shallow water, the shear layer may touch the bottom, in which case
it obviously ceases to grow and Eq.(2.2) applies instead of
Eq.(3.1). At first sight this might seem like a good thing,
since in Eq.(2.2) the drag only increases linearly with
velocity rather than the square of the velocity from
Eq.(3.1). However you have to remember that bottom effects
are felt at low velocities rather than high velocities (since the shear layer
has more time to grow downwards at low speeds) and the point where the two
become equal is where the shear layer separates from the bottom.
So at low velocities the shallow water resistance is linear
(shown by the red line in Fig 6.1)
and greater than would be expected from the quadratic regime
(shown by the blue line)
with an unbounded shear layer. Changing the depth of the
water has the effect of reducing the slope of the linear regime
(differentiating Eq.(2.2)):
so that the transition (at the point + in the diagram)
from linear to quadratic occurs at lower velocities.
So how shallow does the water have to be before you notice the bottom?
You can get an idea of the depth of the shear layer by observing the extent
of the sideways turbulence at the stern of the boat (the shear layer probably
grows downwards at much the same rate as outwards), i.e. around 1 metre.
Any deeper and you shouldn't notice the bottom at all.
The minimum depth for Olympic Regatta courses is, I believe, 2 metres, just to
be on the safe side.
Of course, as the speed tends to zero, the shear layer will
extend to infinity so the actual depth of the water will always be 'noticed'
eventually, but usually at such low resistances that it will be impossible to
distinguish between the linear and quadratic regimes.
Note that moving water will also have flowedinduced
shear (section 5), an entirely separate effect. In that
case (i.e. rivers) the total depth will always be significant.
7. Upstream/Downstream times
It is a common misconception that if you row, say, 2000m upstream, and 2000m
downstream, measured against some fixed points on the bank, your average time
is the same as if you rowed 2000m in still water (I'm ignoring any change in
speed due to fatigue or the effects discussed in the previous sections).
For low stream speeds it is a reasonable approximation,
but the average in stream will always be slower than your still water
time. Why? Because...
Suppose your intrinsic speed through the water is V, the stream
speed is U, and you're rowing a distance L measured along the
bank.
(7.1)
 Still water Time,
 t_{S} = L /V

(7.2)
 Upstream Time,
 t_{U} = L /(VU)

(7.3)
 Downstream Time,
 t_{D} = L /(V+U)

(7.4)
 Average Time,
 t_{A} = ½(t_{U}
+t_{D}) = L.V /(V^{2}
U^{2})

As you'd expect, as stream speed U tends to zero, then the average
of your (upstream+downstream) times tends towards your still
water time, but for any nonzero stream, t_{A} is longer than
t_{S}
(because V^{2}U^{2} is always less than
V^{2}).
How different? Take V=5 m/s (corresponding to t_{S}=6:40
= 400s, for 2000m in still
water). Rowing in a (slowish) stream of U=10 cm/s, you'd row 2000m
upstream in t_{U}=408.2s,
and downstream in t_{D}=392.2s, giving an average of
t_{A}=400.2s, i.e. only 0.2s
slower, not important. But in a quicker stream of U=1m/s, you'd get
t_{A}=416.7s, i.e.
16.7s out, or you'd think you were about 5 lengths slower than you really are.
Note that although the stream speed increased by a factor 10, the error
increased by a factor 100 (depends on U^{2}).