A Question of Unity – Calculating the Velocity Coefficient

I was recently asked to explain why I recommend starting from a value of 1 (i.e. unity) as a first guess for the velocity coefficient, which is guidance based on my conjecture that the “true” value will tend toward 1 for unconstrained alluvial channels in a steady state condition. This probably does deserve an explanation (or 4) because I’ve never heard anyone else recommend the unity conjecture, in spite of the inherent elegance of such simplicity.

The velocity coefficient can be calculated as the square root of slope divided by roughness.

Conventional wisdom dictates that you survey a long-section of the river to obtain slope and then look up roughness from a guide such as Cowan (1956), Chow (1959), or Hicks and Mason (1991). But long-section surveys are never easy, sometimes dangerous, other times impossible, and are error prone (e.g. how long does the section need to be to be long enough to be accurate but short enough that you don’t introduce extraneous error?).

On the other hand, the unity conjecture is easy, cheap, and safe even though it is also unlikely to be exactly correct.

I want to avoid technical jargon or math in this post but the minimum necessary to explain why this matters is that the ratio of the square root of slope over channel roughness is the coefficient needed to solve for velocity using the Manning equation. For reference, the relevant equations are provided in box 1 below. Velocity is the product of this coefficient multiplied by hydraulic radius raised to the power 2/3. Hydraulic radius is a measure of depth, which is obtained by dividing cross-sectional area by wetted perimeter. This is a more accurate indicator of the energy in the stream than the simpler method of calculating depth by dividing area by width. Wetted perimeter will always be longer than width so hydraulic radius will always be less than depth. However, for simplicity and to avoid sounding pretentious, I will only be referring to depth for the rest of this post but just know that I really mean hydraulic radius.

Why does it matter?

Stage-discharge relationships are the product of a stage-area relationship times a stage-velocity relationship. We can obtain a stage-area relationship from a cross-section survey, so that leaves the coefficient in the stage-velocity relationship as the only unexplained variable needed to produce a stage-discharge relationship. The better we can estimate what the stage-velocity relationship “should” look like, the better we can understand and evaluate rating curves. Depth is conveniently related to stage, and stage is conveniently easy to monitor continuously. Within any uniform segment of a stream channel, we can find an offset value such that stage minus this offset is approximately equal to depth. Both area and velocity can then be represented as power law functions based on stage.

Empirical Explanation

I first started exploring the value of the velocity coefficient many years ago by taking data from Hicks and Mason (1991) for many locations and calculating the velocity coefficient. Few stations were exactly 1 but the average from many locations was very close to 1. It was this result from which I first formed the unity conjecture.

Intuitive Explanation

If the coefficient is 1 then velocity must be 1 when depth is 1 (because 1 raised to the power 2/3 is 1). This just seems right. In my field experience, I am sure I measured velocities at a depth of 1 m that could be anywhere from about 0.1 to over 2 m/s. However, this is similar to the spread in the values I found in my Hicks and Mason review that converged around unity. This doesn’t prove or disprove the unity conjecture, it just means that if you have relevant field experience, i.e. you have a good idea of what the actual velocity is at a depth of 1 m, then you can substantially improve on 1 as a first guess for the velocity coefficient.

Logical Explanation

The velocity coefficient will be 1 when the driving force, square root of slope, is exactly equal to the frictional resistance, which is a result predicted by Newton’s third law. If water behaved like a solid block of concrete, then Newton’s third law would mean there could be no velocity because the friction at the base of the block will exactly cancel the force applied to the block.

Fortunately, water is a fluid so it behaves differently.

Consider a barrel of water exactly a meter deep. Suppose that the water is constantly replenished at the rate it discharges from holes cut in the side of the barrel. The velocity at any height will be proportional to the height of the water column above it. Hence the slowest velocity will be at the top of the barrel and the fastest velocity will be from the bottom. It is this principle of pressure causing velocity that Galileo used to predict streamflow when he dabbled in hydraulic engineering. Everyone with practical experience knew that he was wrong, that the fastest velocities were at the surface, not at the streambed, but they couldn’t convince him otherwise. Pitot eventually invented the pitot tube to prove him wrong experimentally. Apparently, calculating stream flow is more complicated than calculating the motion of heavenly bodies.

In a river, frictional resistance is strongest at the bed surface where it exactly cancels out the maximum pressure resulting in a velocity of exactly zero at the bed surface. As you go up in the water column, the effect of friction becomes proportionally less (transference of resistance up the water column is by virtue of water viscosity) with respect to the force of gravity, which results in a vertical velocity profile where point velocities increase to the 1/6th power of depth. When the mean velocity in the vertical is calculated, it turns out the mean velocity increases to the ⅔ power of depth, which is why this exponent is used in the Manning equation.

It also means that when the unity conjecture is true, the stream is in steady state.

If slope increases or decreases relative to resistance, i.e. the velocity coefficient is greater than, or less than, unity (e.g. as happens during a flood wave causing a condition known as discharge hysteresis) then the water will accelerate or decelerate, which is an unsteady state condition.

This brings me to the first condition I place on the conjecture, that the velocity coefficient will be 1 for unconstrained alluvial channels. If slope is high relative to the particle size distribution of the streambed then there is a surplus of energy in the channel that will be used to mobilize sediments causing degradation until either a rougher bed material is exposed or the slope is reduced by either incision or meandering. Conversely, if the slope is small relative to the roughness then the velocity will be low which means that sediments carried from upstream will be deposited resulting in either reduction of roughness by deposition of fine sediments or the channel aggrades, increasing the local slope. An unconstrained alluvial channel is one that can move sediments freely to adjust its width, depth, slope, and roughness, as needed, to comply with Newton’s 3rd law.

Dimensional Analysis Explanation

That water is the unifying element for the SI system of units is very useful. A meter is exactly equal to 1 side of a cube containing exactly 1000 kg of water (by definition). This means that mass to the power ⅓ can readily be converted to a dimension of length and vice versa. This allows us to use Newton’s second law for a zero dimensional velocity model (i.e. a model that is valid for a discrete monitoring point, as is needed for a rating curve, and needless to say a point location can have no mass, which is needed for Newton’s 2nd law).

Velocity has dimensions of Length divided by Time, whereas the right side of Manning equation has the velocity coefficient times Length raised to the ⅔ power. To balance the dimensions, we can infer that the units of the velocity coefficient must be Length to the ⅓ power divided by Time (bold capitalizations are for identities in dimensional analysis). The ⅓ power tells us that the coefficient is the length of one side of a cube that must be orthogonal to depth and this orthogonal length is divided by time, which is exactly what we want – a velocity in the downstream direction.

We would like to multiply this velocity by area to get volumetric flow but for this to be a valid cube (a condition necessary for Newton’s second law to be valid for a zero dimensional model) then the velocity coefficient must be equal to the dimensions of area when area is at unity (i.e. width of 1 times depth of 1 equals area of 1 times velocity of 1). This means that the velocity coefficient must be 1 when the depth is 1 and seeing as the coefficient is a constant, it is therefore 1 at every depth.

I don’t expect anyone to fully agree with all of 4 of these explanations on first read. And I don’t expect anyone to give this a second read unless they are deeply interested in the science and math of rating curves.

My hope is that even casual readers of this post will learn to pay more respect to the control features at their gauges. Those features represent a very clever functional geometry that can be better understood by simplifying the underlying physics. A reasonable first guess of the velocity coefficient may be helpful to understand the shape of a rating curve but it doesn’t replace the need for gaugings. The simplifying assumptions must be tested against actual data. However, with well conceived segmentation of the curve and appropriate adjustment of the offset(s), the assumptions can be very nearly true. 

Readers are invited to disprove the unity conjecture.

My failed attempts at disproving the unity conjecture are the substance of this post. Responses are also welcome for clarifications and corrections needed for any confusion caused by these explanations.


Cowan, W. L. 1956. Estimating hydraulic roughness coefficients. Agricultural Engineering, vol. 37, n°7, p. 473–475.

Chow, V.T. 1959. Open Channel Hydraulics. Blackburn Press.  ISBN 1932846182. Pp.680.

Hicks, DM, and P.D. Mason, 1991. Roughness Characteristics of New Zealand Rivers, DSIR Marine and Freshwater, Wellington


Q = A x V                            where Q is discharge, A is area and V is velocity

h = (stage – offset)

A = C1 x ha                         where C1 is a coefficient and a is an exponent representing the stage-area relation

V = C2 x hb                        where C2 is a coefficient and b is an exponent representing the stage-velocity relation

=> Q = C1 x C2 x ha+b

V = S/n x R2/3              Manning equation where S is slope, n is the roughness factor and  R is hydraulic radius, which is a measure of channel depth 

R = A/Pw,                      Hydraulic radius where Pw is the wetted perimeter of the channel

The Unity Conjecture:

S/n = 1                          for unconstrained alluvial channels in steady state conditions

V = R = h                        for h = 1

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