---
title: Side Overflow Weir Derivations
description: "# A side-overflow weir A side-overflow weir is a deliberately lowered length of wall along the side of a drain or channel. In a side-overflow weir, the top of that lowered wall is the **crest**. When the channel water is…"
url: "https://mes100.com/docs/side-overflow-weir-derivations/"
updated: "2026-11-12"
category: Exposition
---

# Side Overflow Weir Derivations

# A side-overflow weir

A side-overflow weir is a deliberately lowered length of wall along the side of a drain or channel. In a side-overflow weir, the top of that lowered wall is the **crest**. When the channel water is below the crest, it continues along the drain. When the water rises above the crest, part of the flow can spill sideways into a receiving channel or an off-line detention pond.

An ordinary transverse weir, on the other hand, has its crest across the flow path, so all the water crosses the crest on its way downstream. [Figure 1](#fig-side-weir-comparison) shows that difference in plan.

Figure 1. A transverse weir crosses the flow path; a side weir runs beside it

Suppose a flood flow of 5 m³/s arrives along a drain, but the drain downstream can safely carry only 3 m³/s. A side weir must divert the remaining 2 m³/s into a pond. The design question is then: at what height should we set the crest, and how long must the crest be to spill those 2 m³/s?

[Figure 2](#fig-side-weir-design-flow) shows the flow split we are designing for.

Figure 2. Divert 2 m³/s so that 3 m³/s continues downstream

The flow spilling over each metre of crest depends on how high the water stands above the crest at that point. As water leaves the drain over the weir, this head changes along the opening, so each metre of crest spills a different amount. The total diverted flow is the sum of these spills along the whole crest. Our task is therefore to choose a crest height and length whose spills add up to 2 m³/s.

When we check an existing weir instead, the crest height and length are already fixed. We then calculate how much flow the weir diverts for the given drain and pond water levels.

In this article, we will first derive a general crest-length equation from continuity. We will then use [Equation (6)](#eq-side-weir-general-length) to derive the MSMA formulas for two conditions: when the water surface rises along the weir, and when it falls.

# How the flow divides along the crest

Figure 3. A side weir divides the incoming flow between the channel and a pond

In [Figure 3](#fig-side-weir-plan), the channel has constant width B  in metres. Section 1 is the upstream end of the crest and section 2 its downstream end. The incoming discharge is Q_1 , the continuing channel discharge is Q_2 , and the total sideways discharge is Q_s , all in cubic metres per second. For steady flow with no other inflow or outlet,

 Qs=Q1−Q2.Q_s=Q_1-Q_2.(1)

The [ordinary-weir derivation](https://mes100.com/docs/weir-discharge-from-bernoulli/) explains how discharge depends on head above a crest. A side weir differs because water leaves gradually along its length, so both the channel discharge and the head above the crest change from one end of the crest to the other.

A **water-surface profile** traces the water level along the channel. Water at rest has a level surface, while flowing water can have a sloping surface. For a steady reach with equal speeds at its two ends, the upstream water surface must be higher by the head lost along the reach. In uniform flow along a sloping channel, the water surface falls with the bed while depth stays constant. The [tailwater and backwater article](https://mes100.com/docs/tailwater-backwater-from-bernoulli/) explains this energy balance and shows how water level and local depth can change by different amounts.

[Figure 4](#fig-side-weir-route) shows how the article is organised. We first derive the equations common to both cases, then two ways of relating water depth to the remaining discharge. The last two sections apply these to the rising and falling profiles, using MSMA’s coefficients and assumptions.

Figure 4. From the spatial-flow method to the two MSMA formulas

# Analytical framework for side overflow

In this section we build the equations from three parts: continuity, an overflow law for each metre of crest, and an energy relation between depth and discharge. MSMA prints only the final formulas; the steps here show where they come from. MSMA’s own coefficients and limits come in the later sections.

## Geometry and notation

Figure 5. Channel depth and overflow head use different reference levels

We model the straight, rectangular channel in [Figure 5](#fig-side-weir-section) with a level bed and a horizontal crest along one side. The channel depth y  and crest height c  are measured from the invert, so the overflow head is h=y-c . All three are in metres. With both the bed and crest level, a rise in water level increases depth and overflow head by the same amount. We use lowercase c  for crest height because MSMA 2 also uses uppercase C  for the dimensionless coefficient in the rising-profile equation.

The remaining channel discharge at any section is Q , in cubic metres per second. Flow is steady, and the crest has no other inflow or outlet besides the side overflow.

## Continuity and overflow along a short crest strip

Figure 6. The water leaving each strip reduces the continuing discharge

Let x  be distance downstream from the start of the crest, so x=0  at section 1 and x=L  at section 2. The small length dx  in [Figure 6](#fig-side-weir-strip) loses water sideways at a rate q_\\ell\\,dx . Here q_\\ell  is the positive lateral discharge per metre of crest, in \\mathrm{m^2/s} . Subtracting the downstream discharge from the upstream discharge gives

 Q−(Q+dQ)=qℓ dx,dQdx=−qℓ.Q-(Q+dQ)=q_\\ell\\,dx,\\qquad \\frac{dQ}{dx}=-q_\\ell.(2)

For a freely discharging, sharp crest, we represent the local overflow by the following law, where g=9.81\\ \\mathrm{m/s^2}  is gravitational acceleration:

 qℓ=k2g (y−c)3/2,y>c.q_\\ell=k\\sqrt{2g}\\,(y-c)^{3/2},\\qquad y>c.(3)

The dimensionless coefficient k  belongs to this particular form of the equation, and its value must be calibrated from measurements for the side-weir geometry and flow conditions. If we write the same law in the more familiar weir form q_\\ell=(2/3)C_d\\sqrt{2g}\\,h^{3/2} , where C_d  is a discharge coefficient, then k=(2/3)C_d . It should be noted that the factor 2/3 must be carried over whenever we convert a coefficient from one form to the other.

When y\\leq c , that strip has no free overflow, and we set its lateral discharge to zero.

## The general length equation

We now derive the relation between crest length and the diverted discharge. In design, we use it to find the length that reduces the channel discharge from Q_1  to Q_2 , diverting Q_s=Q_1-Q_2 . For an existing weir, we use the same relation the other way round, together with the channel’s boundary conditions, to find the end heads and the diversion.

The head above the crest can change along its length, so the sideways discharge per metre can also change. We therefore calculate the length needed for each small reduction in channel discharge and add those lengths. To relate the remaining discharge to the local head, we write Q(h) , keeping the geometry and other model parameters fixed. On a smooth segment where head changes monotonically and q_\\ell>0 , the chain rule gives

 dQdx=dQdhdhdx=−qℓ(h).\\frac{dQ}{dx}=\\frac{dQ}{dh}\\frac{dh}{dx}=-q_\\ell(h).(4)

Equivalently, the discharge change over a short crest length is dQ=-q_\\ell\\,dx . Substituting dQ=(dQ/dh)\\,dh  and rearranging gives

 dx=−1qℓ(h)dQdh dh.dx=-\\frac{1}{q_\\ell(h)}\\frac{dQ}{dh}\\,dh.(5)

The required crest length L  is measured in metres along the channel from section 1 to section 2. Let h_1  and h_2  be the heads above the crest at those upstream and downstream ends. [Figure 7](#fig-side-weir-general-length) shows how the small lengths dx  make up L . The illustrated head rises downstream; for a falling profile, we keep the same section numbering and have h_2<h_1 .

Figure 7. Adding the short crest lengths gives the total length L

Each small discharge reduction -dQ  divided by the local side discharge per metre q_\\ell  gives the required short length dx . Adding these lengths from section 1 to section 2, using the corresponding heads as the integration bounds, gives

 L=−∫h1h21qℓ(h)dQdh dh.L=-\\int_{h_1}^{h_2}\\frac{1}{q_\\ell(h)}\\frac{dQ}{dh}\\,dh.(6)

Substituting the free-overflow law, [Equation (3)](#eq-side-weir-strip-overflow), into [Equation (6)](#eq-side-weir-general-length) with a constant coefficient k  gives

 L=−1k2g∫h1h2dQ/dhh3/2 dh.L=-\\frac{1}{k\\sqrt{2g}}\\int_{h_1}^{h_2}\\frac{dQ/dh}{h^{3/2}}\\,dh.(7)

[Equation (7)](#eq-side-weir-general-length-strip) is the starting equation for both cases. We integrate from the upstream head to the downstream head along the actual profile; the leading minus sign makes the length positive because the discharge decreases downstream. To evaluate the integral, we need an expression for Q(h)  that is smooth along the crest. If the profile has a hydraulic jump, switches between subcritical and supercritical flow, or rises and then falls, we split the crest into segments or integrate in x  instead.

## Two energy approximations

To calculate the crest length, we need to relate water depth to the flow remaining in the drain. We use two models, and they differ only in how they write the energy that stays constant along the crest:

 E=c+h+V22g⏟constant specific energyEw=αfV22g+ηh⏟constant corrected head\\underbrace{E=c+h+\\frac{V^2}{2g}}_{\\text{constant specific energy}}\\qquad\\qquad\\underbrace{E_w=\\alpha_f\\frac{V^2}{2g}+\\eta h}_{\\text{constant corrected head}}(8)

Here V  is the mean channel speed, E  is measured above the invert and E_w  above the crest. The factors \\alpha_f  and \\eta  correct the velocity head and pressure head; we explain them with the [second model](#corrected-head-model) below. With \\alpha_f=\\eta=1 , the two models are the same, since then E_w=E-c . Continuity, the overflow law and the length integral are shared by both.

[Figure 8](#fig-side-weir-settings) shows two side-overflow arrangements that we will examine.

Figure 8. The remaining water slows and deepens, or speeds up and becomes shallower

In both arrangements, water spills sideways into the pond. They differ in what happens to the water that continues along the drain.

In the left-hand example, the water level farther downstream holds the flow up. Along the side opening, the remaining water slows and becomes deeper, even though less water continues downstream.

In the right-hand example, the low side crest and downstream conditions allow the remaining water to run on at a lower level. It speeds up and becomes shallower along the opening. MSMA’s low-crest, falling-profile formula concerns this second arrangement.

Either arrangement can occur on a gently sloping bed. A drain’s slope alone does not tell us how its water surface will change. The crest height and the water level farther downstream also matter.

The formulas in this article leave out other conditions, such as a high side crest, a pond drowning the crest, downstream backwater, significant losses, a critical-flow transition or hydraulic jump, or rapidly changing water levels. These may need a full water-surface profile calculation or additional treatment.

### Constant specific energy with hydrostatic pressure

#### Assumptions and head relation

We keep the level-bed rectangular channel and assume hydrostatic pressure, with kinetic-energy correction \\alpha=1 , as if the speed were uniform across the section. The mean speed is V=Q/(By) , so the specific energy E , in metres above the invert, is

 E=y+V22g=y+Q22gB2y2.E=y+\\frac{V^2}{2g}=y+\\frac{Q^2}{2gB^2y^2}.(9)

Hydrostatic pressure means that pressure at a point comes from the weight of water above it, with negligible vertical acceleration, as assumed in the [USACE channel-flow equations](https://www.hec.usace.army.mil/confluence/hmsdocs/hmstrm/channel-flow/channel-flow-basic-concepts-equations-and-solution-techniques). Combining this pressure relation with the [energy equation](https://mes100.com/docs/from-newtons-second-law-to-bernoulli/) gives the expression above.

We hold E  constant along the crest. This neglects friction losses along the channel, and assumes that the water leaving sideways leaves the specific energy of the water staying in the channel unchanged.

#### Discharge and surface profile

Rearranging [Equation (9)](#eq-side-weir-specific-energy) gives the channel discharge at each depth:

 Q(y)=By2g(E−y),c<y<E.Q(y)=By\\sqrt{2g(E-y)},\\qquad c<y<E.(10)

We choose the positive square root because the main flow continues downstream. Differentiating this expression with respect to depth gives

 dQdy=B2g E−3y/2E−y.\\frac{dQ}{dy}=B\\sqrt{2g}\\,\\frac{E-3y/2}{\\sqrt{E-y}}.(11)

For a rectangular section, the dimensionless Froude number is Fr=V/\\sqrt{gy} . Subcritical flow has Fr<1 , and supercritical flow has Fr>1 . The two profiles in [Figure 9](#fig-side-weir-branches) follow from the same depth-gradient equation.

Figure 9. Constant energy allows a rising or falling depth as discharge decreases

We differentiate [Equation (9)](#eq-side-weir-specific-energy) along the channel. Both Q  and y  vary with x , while B  and g  are constant:

 0=dEdx=dydx+QgB2y2dQdx−Q2gB2y3dydx.0=\\frac{dE}{dx}=\\frac{dy}{dx}+\\frac{Q}{gB^2y^2}\\frac{dQ}{dx}-\\frac{Q^2}{gB^2y^3}\\frac{dy}{dx}.(12)

Collect the two depth-gradient terms in [Equation (12)](#eq-side-weir-energy-gradient) and substitute dQ/dx=-q_\\ell  from [Equation (2)](#eq-side-weir-strip-continuity):

 (1−Q2gB2y3)dydx=QqℓgB2y2.\\left(1-\\frac{Q^2}{gB^2y^3}\\right)\\frac{dy}{dx}=\\frac{Qq_\\ell}{gB^2y^2}.(13)

Substituting V=Q/(By)  into the Froude-number definition gives Fr^2=Q^2/(gB^2y^3) . We obtain

 dydx=QqℓgB2y2(1−Fr2).\\frac{dy}{dx}=\\frac{Qq_\\ell}{gB^2y^2(1-Fr^2)}.(14)

This expression rests on four assumptions together: hydrostatic pressure, constant specific energy, \\alpha=1  and a rectangular channel of constant width. Under them, it covers both rising and falling profiles. The corrected-head model later in the article uses a different discharge relation, so this equation does not carry over to it.

[Equation (14)](#eq-side-weir-surface-gradient) tells us how depth changes as we move downstream: dy/dx>0  means rising depth, while dy/dx<0  means falling depth. For forward flow and positive side overflow, the numerator is positive. The sign of 1-Fr^2  therefore gives the two profiles in [Figure 9](#fig-side-weir-branches):

- With subcritical flow, Fr<1 , depth increases downstream. Since V^2/(2g)=E-y , the velocity head decreases as the water deepens.
- With supercritical flow, Fr>1 , depth decreases downstream. The velocity head increases even though less discharge remains in the channel.

At Fr=1 , the coefficient of dy/dx  is zero while the right-hand side stays positive, so no finite slope satisfies the equation. A smooth profile therefore cannot pass through critical flow. A critical entrance, a transition between subcritical and supercritical flow, or a hydraulic jump needs its own calculation.

#### Crest-length relation

Because h=y-c  and c  is constant, dh=dy  and dQ/dh=dQ/dy . Substitute the discharge derivative from [Equation (11)](#eq-side-weir-discharge-depth-gradient) and the overflow law from [Equation (3)](#eq-side-weir-strip-overflow) into [Equation (5)](#eq-side-weir-length-differential). Cancelling \\sqrt{2g}  gives

 dxdy=B2k 3y−2EE−y(y−c)3/2.\\frac{dx}{dy}=\\frac{B}{2k}\\,\\frac{3y-2E}{\\sqrt{E-y}(y-c)^{3/2}}.(15)

Let y_1  and y_2  be the upstream and downstream depths. Integrating between them gives

 L=B2k∫y1y23y−2EE−y(y−c)3/2 dy.L=\\frac{B}{2k}\\int_{y_1}^{y_2}\\frac{3y-2E}{\\sqrt{E-y}(y-c)^{3/2}}\\,dy.(16)

The same integral applies to a falling profile that stays supercritical throughout: the integrand and the depth increment are then both negative, so the length is still positive. MSMA’s falling-profile formula, discussed later, uses a different energy model.

The two positive heads under the square roots suggest a substitution. [Figure 10](#fig-side-weir-integration-heads) shows the velocity head E-y  between the energy line and water surface, and the overflow head y-c  between the surface and crest.

Figure 10. Specific energy combines water depth and velocity head

We introduce the dimensionless integration variable t=\\sqrt{(E-y)/(y-c)} . It is the square root of the ratio of velocity head to overflow head. Rearranging its definition gives

 y=E+ct21+t2,dydt=−2(E−c)t(1+t2)2.y=\\frac{E+ct^2}{1+t^2},\\qquad \\frac{dy}{dt}=-\\frac{2(E-c)t}{(1+t^2)^2}.(17)

Since E>y>c , both square roots are real and t>0 . Substituting y  and dy/dt  from [Equation (17)](#eq-side-weir-substitution-depth) into [Equation (15)](#eq-side-weir-depth-length-gradient) gives

 dxdt=−BkE+(3c−2E)t2(E−c)(1+t2)=Bk\[2E−3cE−c−31+t2\].\\frac{dx}{dt}=-\\frac{B}{k}\\frac{E+(3c-2E)t^2}{(E-c)(1+t^2)}=\\frac{B}{k}\\left\[\\frac{2E-3c}{E-c}-\\frac{3}{1+t^2}\\right\].(18)

At the labelled end sections, t_1=\\sqrt{(E-y_1)/(y_1-c)}  and t_2=\\sqrt{(E-y_2)/(y_2-c)} . We evaluate the length between those same sections:

 L=Bk\[2E−3cE−ct−3tan⁡−1t\]t1t2.L=\\frac{B}{k}\\left\[\\frac{2E-3c}{E-c}t-3\\tan^{-1}t\\right\]_{t_1}^{t_2}.(19)

The two terms integrate to a linear term and an arctangent. Define our dimensionless flow function \\Psi  by

 Ψ(y;E,c)=2E−3cE−cE−yy−c−3tan⁡−1E−yy−c.\\Psi(y;E,c)=\\frac{2E-3c}{E-c}\\sqrt{\\frac{E-y}{y-c}}-3\\tan^{-1}\\sqrt{\\frac{E-y}{y-c}}.(20)

The arctangent is in radians. Evaluating at the physical start and end sections gives

 L=Bk\[Ψ(y2;E,c)−Ψ(y1;E,c)\].L=\\frac{B}{k}\\left\[\\Psi(y_2;E,c)-\\Psi(y_1;E,c)\\right\].(21)

\\Psi  itself can be negative, but the difference \\Psi(y_2)-\\Psi(y_1)  is positive on either branch when y_1  is the upstream depth and y_2  the downstream depth. Only the difference matters, so a chart may shift \\Psi  by a constant without changing the length.

To display the dimensionless geometry, define u=y/E  and r=c/E . Then the flow function in [Equation (20)](#eq-side-weir-de-marchi-function) can be written as

 Ψ(u;r)=2−3r1−r1−uu−r−3tan⁡−11−uu−r,r<u<1.\\Psi(u;r)=\\frac{2-3r}{1-r}\\sqrt{\\frac{1-u}{u-r}}-3\\tan^{-1}\\sqrt{\\frac{1-u}{u-r}},\\qquad r<u<1.(22)

Note that \\Psi  depends on the crest ratio r  as well as on y/E . A handbook may write the flow function as a function of y/E  alone, but any chart or table for it must still allow for the crest height.

### Constant corrected head with velocity and pressure factors

#### Physical picture

The constant-energy model above takes the velocity head and pressure head at face value. For a low crest with a falling profile, the handbook model behind MSMA’s formula instead weights each term with a dimensionless correction factor: \\alpha_f  for the velocity head and \\eta  for the pressure head. Near the side opening, water accelerates sideways and downwards towards the crest, so the pressure there is lower than hydrostatic and the pressure head is less than the overflow head h ; a value of \\eta  below 1 represents that reduction. The suffix f  marks this falling-profile model. We write the resulting corrected head as E_w , in metres above the crest; the subscript w  refers to the weir crest. [Figure 11](#fig-side-weir-falling-strip) shows the two weighted terms.

Figure 11. The corrected model weights velocity and pressure head separately

#### Assumptions and head relation

We hold E_w  and both correction factors constant along the crest. Their values, and the heads at the two ends of the crest, stay open for now; the falling-profile section later supplies them.

In [Figure 11](#fig-side-weir-falling-strip), the depth is c+h  and the full channel area is B(c+h) . The crest-based head relation is

 Ew=αfV22g+ηh.E_w=\\alpha_f\\frac{V^2}{2g}+\\eta h.(23)

#### Discharge and surface profile

We take \\alpha_f>0 , \\eta>0 , E_w>0  and g>0 . On the overflowing segment, h>0  and E_w-\\eta h>0 , so the square root gives a positive forward speed and the derivative below has a nonzero denominator. Solving the head relation for V  and multiplying by the full channel area gives

 V(h)=2gαf(Ew−ηh),Q(h)=B(c+h)2gαf(Ew−ηh).V(h)=\\sqrt{\\frac{2g}{\\alpha_f}(E_w-\\eta h)},\\qquad Q(h)=B(c+h)\\sqrt{\\frac{2g}{\\alpha_f}(E_w-\\eta h)}.(24)

With E_w , \\alpha_f  and \\eta  constant along the crest, differentiating the product gives

 dQdh=B2gαf\[Ew−ηh−η(c+h)2Ew−ηh\]\\frac{dQ}{dh}=B\\sqrt{\\frac{2g}{\\alpha_f}}\\left\[\\sqrt{E_w-\\eta h}-\\frac{\\eta(c+h)}{2\\sqrt{E_w-\\eta h}}\\right\](25) dQdh=B2gαfEw−ηc/2−3ηh/2Ew−ηh.\\hphantom{\\frac{dQ}{dh}}=B\\sqrt{\\frac{2g}{\\alpha_f}}\\frac{E_w-\\eta c/2-3\\eta h/2}{\\sqrt{E_w-\\eta h}}.(26)

Continuity gives dh/dx=-q_\\ell/(dQ/dh) . Thus when dQ/dh  is positive the head falls downstream, and when it is negative the head rises. The dividing line is dQ/dh=0 . With both correction factors equal to 1 this is the familiar Fr=1 , but with the handbook’s factors it falls elsewhere.

#### Crest-length relation

We now substitute the discharge derivative from [Equation (26)](#eq-side-weir-corrected-discharge-gradient-simplified) into [Equation (5)](#eq-side-weir-length-differential), with the same local overflow law from [Equation (3)](#eq-side-weir-strip-overflow). Along a falling segment with h_2<h_1  and dQ/dh>0 , this gives

 dx=−BkαfEw−ηc/2−3ηh/2Ew−ηh h3/2 dh.dx=-\\frac{B}{k\\sqrt{\\alpha_f}}\\frac{E_w-\\eta c/2-3\\eta h/2}{\\sqrt{E_w-\\eta h}\\,h^{3/2}}\\,dh.(27)

The minus sign converts a decreasing head into a positive downstream length. Integrating from section 1 to section 2, then reversing the head bounds, gives

 L=Bkαf∫h2h1Ew−ηc/2−3ηh/2Ew−ηh h3/2 dh.L=\\frac{B}{k\\sqrt{\\alpha_f}}\\int_{h_2}^{h_1}\\frac{E_w-\\eta c/2-3\\eta h/2}{\\sqrt{E_w-\\eta h}\\,h^{3/2}}\\,dh.(28)

# When the water surface rises along the weir

MSMA illustrates this condition with subcritical flow in a mild channel. As water spills sideways, the continuing flow slows and deepens, as shown in [Figure 12](#fig-side-weir-length).

Figure 12. The required diversion fixes the remaining flow at the end of the crest

## MSMA’s rising-profile formula

For this condition we take the subcritical branch, Fr<1 , of [Equation (14)](#eq-side-weir-surface-gradient), so that dy/dx>0 . The assumptions of hydrostatic pressure, \\alpha=1  and constant specific energy all carry over, and the length is the flow-function difference in [Equation (21)](#eq-side-weir-de-marchi-length).

MSMA 2 Eq. 20.13 gives the following expression.

 L=BCm\[ϕ ⁣(y2E)−ϕ ⁣(y1E)\].L=\\frac{B}{C_m}\\left\[\\phi\\!\\left(\\frac{y_2}{E}\\right)-\\phi\\!\\left(\\frac{y_1}{E}\\right)\\right\].(29)

Here y_1  and y_2  are the endpoint depths, E  is specific energy above the invert, and \\phi  is the flow function attributed to Collinge (1957). We write MSMA’s coefficient as C_m  to distinguish it from crest height. MSMA 2 gives C_m=0.35  for a free nappe and a best-working Froude-number range of 0.3–0.92.

## Match the derived function to MSMA

MSMA names \\phi  but gives neither its equation nor a chart. We can, however, check our \\Psi  against another source. Bos’s [*Discharge Measurement Structures*](https://edepot.wur.nl/64285), third revised edition (1989), Annex 3, Eq. A3.7 and Figure A3.3, give the same De Marchi bracket derived above, written with an arcsine. The identity

 tan⁡−1E−yy−c=sin⁡−1E−yE−c,c<y<E,\\tan^{-1}\\sqrt{\\frac{E-y}{y-c}}=\\sin^{-1}\\sqrt{\\frac{E-y}{E-c}},\\qquad c<y<E,(30)

shows that our arctangent and that arcsine are the same angle. Bos uses an effective coefficient C_s  with a length multiplier 3^{3/2}B/(2C_s) . Setting k=2C_s/3^{3/2}  gives exactly our B/k  multiplier. Whenever we take a flow function from one source, we must take its coefficient convention with it.

For comparison with MSMA, the required normalization is

 ϕ2−ϕ1Cm=Ψ2−Ψ1k.\\frac{\\phi_2-\\phi_1}{C_m}=\\frac{\\Psi_2-\\Psi_1}{k}.(31)

Here subscripts 1 and 2 mean evaluation at the two end sections. With the standard bracket \\phi=\\Psi , MSMA’s form is recovered by C_m=k . Its value 0.35 then means q_\\ell=0.35\\sqrt{2g}\\,h^{3/2} , equivalent to C_d=0.525  in the alternative strip convention or C_s=0.90933  in Bos’s convention. The 0.35 comes from measurement; the continuity and energy equations then carry it along the crest to give the length.

We can therefore evaluate [Equation (29)](#eq-side-weir-msma-rising) with \\phi=\\Psi  and C_m=k . The calculation below uses our own definitions of \\Psi  and k .

## Calculate a rising-profile length and diversion

Figure 13. The worked example loses discharge while its depth rises

We specify an illustrative channel width B=2.00\\ \\mathrm{m} , crest height c=0.70\\ \\mathrm{m} , constant specific energy E=1.80\\ \\mathrm{m} , upstream depth y_1=1.30\\ \\mathrm{m}  and downstream depth y_2=1.70\\ \\mathrm{m} , as shown in [Figure 13](#fig-side-weir-example). We simply assume these end depths here. In an actual design, the end depths come from the connecting channel or its control.

For this example, we take C_d=0.60  in the strip-weir form above, giving k=(2/3)(0.60)=0.40 . This is an illustrative value, not a recommended coefficient; we compare it with MSMA’s C_m=0.35  below.

First we calculate the channel discharges using [Equation (10)](#eq-side-weir-hydrostatic-discharge):

 Q1=2(1.30)2(9.81)(1.80−1.30)=8.14344 m3/s,Q_1=2(1.30)\\sqrt{2(9.81)(1.80-1.30)}=8.14344\\ \\mathrm{m^3/s},(32) Q2=2(1.70)2(9.81)(1.80−1.70)=4.76243 m3/s.Q_2=2(1.70)\\sqrt{2(9.81)(1.80-1.70)}=4.76243\\ \\mathrm{m^3/s}.(33)

The corresponding Froude numbers are

 Fr1=2(1.80−1.30)1.30=0.8771,Fr2=2(1.80−1.70)1.70=0.3430.Fr_1=\\sqrt{\\frac{2(1.80-1.30)}{1.30}}=0.8771,\\qquad Fr_2=\\sqrt{\\frac{2(1.80-1.70)}{1.70}}=0.3430.(34)

Both endpoints are subcritical. Since Fr^2=2(E-y)/y  decreases as depth increases, the intermediate sections also stay within 0.3430–0.8771. This lies within the 0.3–0.92 range that MSMA 2 gives for the rising-profile formula.

Using the substitution in [Equation (17)](#eq-side-weir-substitution-depth) and the flow function in [Equation (20)](#eq-side-weir-de-marchi-function), the endpoint values are

 t1=0.500.60=0.912871,t2=0.101.00=0.316228,t_1=\\sqrt{\\frac{0.50}{0.60}}=0.912871,\\qquad t_2=\\sqrt{\\frac{0.10}{1.00}}=0.316228,(35) Ψ1=−0.97481833,Ψ2=−0.48761243.\\Psi_1=-0.97481833,\\qquad \\Psi_2=-0.48761243.(36)

Substituting these flow-function values into [Equation (21)](#eq-side-weir-de-marchi-length) gives the crest length:

 L=2.000.40\[−0.48761243−(−0.97481833)\]=2.43603 m.L=\\frac{2.00}{0.40}\\left\[-0.48761243-(-0.97481833)\\right\]=2.43603\\ \\mathrm{m}.(37)

The coefficient k=0.40  is our own choice for this example. MSMA recommends a lower value, 0.35. With MSMA’s value and the same end depths, the length L_{0.35}  is

 L0.35=2.000.35(0.48720590)=2.78403 m,L_{0.35}=\\frac{2.00}{0.35}(0.48720590)=2.78403\\ \\mathrm{m},(38)

14.3% longer, because each metre of crest spills less water with the lower coefficient. The difference comes only from the coefficient we choose; MSMA’s formula itself agrees with our derivation. For the rest of this example we keep k=0.40  and L=2.43603\\ \\mathrm{m} .

Using the endpoint discharges from [Equation (32)](#eq-side-weir-rising-upstream-discharge) and [Equation (33)](#eq-side-weir-rising-downstream-discharge) in the continuity relation, [Equation (1)](#eq-side-weir-continuity), gives the diverted flow:

 Qs=8.14344−4.76243=3.38101 m3/s.Q_s=8.14344-4.76243=3.38101\\ \\mathrm{m^3/s}.(39)

As a check, we also integrated the spill from each strip numerically along the calculated profile. The total is the same 3.38101 m³/s, so the end discharges and the spill along the crest account for the same water.

### Using one head for the whole crest

It is tempting to calculate the spill with a single head for the whole crest, as a network model connection often does. [Figure 14](#fig-side-weir-one-head-example) compares the actual rising head with each end head applied along the whole crest. The length, geometry and coefficient stay fixed.

Figure 14. One endpoint head gives a different spill estimate for the same crest

If we replace the varying head by its upstream value h_1=1.30-0.70=0.60\\ \\mathrm{m} , the same strip law predicts

 Qs,1=0.402(9.81)(2.43603)(0.60)3/2=2.00594 m3/s.Q_{s,1}=0.40\\sqrt{2(9.81)}(2.43603)(0.60)^{3/2}=2.00594\\ \\mathrm{m^3/s}.(40)

The single-head estimate in [Equation (40)](#eq-side-weir-single-head-diversion) is 40.7% below the integrated result in [Equation (39)](#eq-side-weir-rising-diversion). Using the downstream head h_2=1.00\\ \\mathrm{m}  instead gives 4.31611 m³/s, 27.7% too high. The crest, length and coefficient are the same in all three calculations; only the head differs.

To reproduce the integrated discharge with a single head, we would need an effective head of about 0.84977 m, and that value holds only for this flow state. A network model that uses one head for a side weir therefore needs that head calibrated against the actual profile.

# When the water surface falls along the weir

A low side crest can draw the water surface down along the opening, even in a mildly sloping drain. MSMA also illustrates falling profiles in steep channels. Before using MSMA’s falling-profile formula, we check the low-crest criterion and confirm that the entrance and downstream conditions allow the water surface to fall. [Figure 15](#fig-side-weir-falling-datum) shows the falling head above the crest.

Figure 15. The falling-profile formula measures Ew above the crest

## MSMA’s falling-profile formula

MSMA 2 Eqs. 20.11–20.12 give, with c  replacing their crest-height symbol C ,

 L=2.03B(5.28−2.63cEw),L=2.03B\\left(5.28-2.63\\frac{c}{E_w}\\right),(41) Ew=αVn22g+α′(yn−c).E_w=\\alpha\\frac{V_n^2}{2g}+\\alpha'(y_n-c).(42)

Here y_n  and V_n  are approach normal depth and mean speed. The velocity and pressure-head corrections are \\alpha  and \\alpha' . MSMA 2 gives approach values 1.2 and 1.0, and lower-end values 1.4 and 0.95, respectively.

The head E_w  is measured above the crest, as in [Figure 15](#fig-side-weir-falling-datum). With unit correction factors, E=E_w+c ; using E  measured from the invert in this formula would give the wrong ratio c/E_w .

In the next section we show that the bracket 5.28-2.63c/E_w  follows from the corrected-head model and its assumed end heads, while the multiplier 2.03 also contains a measured coefficient.

## Apply the head corrections and endpoint assumptions

To reconstruct those constants, we use the model reported in APWA/FWQA’s [*Combined Sewer Regulation and Management*](https://nepis.epa.gov/Exe/ZyPURL.cgi?Dockey=400002WS.TXT) (1970), pp. 37–38: free overflow, constant width and crest height, velocity correction 1.4, pressure correction 0.8, and endpoint heads h_1=E_w/2  and h_2=E_w/20 . We apply these values to the corrected-head length relation, [Equation (28)](#eq-side-weir-corrected-length). Note that this pressure correction, 0.8, differs from MSMA’s lower-end value of 0.95; we return to that difference after integrating. The handbook demonstrates the method on a sewer sloping at 0.002, or 1 in 500, so a steep bed is not required.

Define z=h/E_w  and s=c/E_w , both dimensionless. The endpoint values are z_1=h_1/E_w  and z_2=h_2/E_w . [Figure 16](#fig-side-weir-falling-bounds) locates the assumed heads at the two crest ends.

Figure 16. The bracket constants use heads of Ew over two and Ew over twenty

The endpoint heads in [Figure 16](#fig-side-weir-falling-bounds) give

 z1=12,z2=120,η=0.8.z_1=\\frac12,\\qquad z_2=\\frac1{20},\\qquad \\eta=0.8.(43)

With s<0.6 , the numerator 1-\\eta s/2-3\\eta z/2  remains positive over these bounds, as required for a decreasing head to accompany a decreasing channel discharge in this corrected-head model.

Substituting h=E_wz  and dh=E_w\\,dz  into [Equation (28)](#eq-side-weir-corrected-length) cancels every dimensional factor of E_w  in the integrand:

 L=Bkαf∫z2z11−ηs/2−3ηz/21−ηz z3/2 dz.L=\\frac{B}{k\\sqrt{\\alpha_f}}\\int_{z_2}^{z_1}\\frac{1-\\eta s/2-3\\eta z/2}{\\sqrt{1-\\eta z}\\,z^{3/2}}\\,dz.(44)

The numerator is linear in s , which explains why MSMA’s bracket is linear in c/E_w . For \\eta>0  and 0<z_2<z_1<1/\\eta , define two endpoint quantities

 Rj=1zj−η,θj=sin⁡−1ηzj,j=1,2,R_j=\\sqrt{\\frac1{z_j}-\\eta},\\qquad \\theta_j=\\sin^{-1}\\sqrt{\\eta z_j},\\qquad j=1,2,(45)

where j  labels the two crest-end sections and the angles \\theta_j  are in radians. The required primitives are

 ∫dz1−ηz z3/2=−21z−η,\\int\\frac{dz}{\\sqrt{1-\\eta z}\\,z^{3/2}}=-2\\sqrt{\\frac1z-\\eta},(46) ∫dz1−ηz z=2ηsin⁡−1ηz.\\int\\frac{dz}{\\sqrt{1-\\eta z}\\,\\sqrt z}=\\frac{2}{\\sqrt\\eta}\\sin^{-1}\\sqrt{\\eta z}.(47)

Splitting the numerator of [Equation (44)](#eq-side-weir-dimensionless-falling-length) into its constant, s  and z  terms, then using the antiderivatives in [Equation (46)](#eq-side-weir-falling-first-antiderivative) and [Equation (47)](#eq-side-weir-falling-second-antiderivative) between z_2  and z_1 , gives

 L=Bkαf{2(R2−R1)−3η(θ1−θ2)−η(R2−R1)cEw}.L=\\frac{B}{k\\sqrt{\\alpha_f}}\\left\\{2(R_2-R_1)-3\\sqrt\\eta(\\theta_1-\\theta_2)-\\eta(R_2-R_1)\\frac{c}{E_w}\\right\\}.(48)

For \\eta=0.8 , z_1=0.5  and z_2=0.05 ,

 R1=1.2=1.095445,R2=19.2=4.381780,R_1=\\sqrt{1.2}=1.095445,\\qquad R_2=\\sqrt{19.2}=4.381780,(49) θ1=sin⁡−10.4,θ2=sin⁡−1(0.2).\\theta_1=\\sin^{-1}\\sqrt{0.4},\\qquad \\theta_2=\\sin^{-1}(0.2).(50)

The two bracket coefficients are therefore

 2(R2−R1)−30.8(θ1−θ2)=5.27567627≈5.28,2(R_2-R_1)-3\\sqrt{0.8}(\\theta_1-\\theta_2)=5.27567627\\approx5.28,(51) 0.8(R2−R1)=2.62906828≈2.63.0.8(R_2-R_1)=2.62906828\\approx2.63.(52)

The coefficients in [Equation (51)](#eq-side-weir-falling-constant-term) and [Equation (52)](#eq-side-weir-falling-crest-term) round to MSMA’s 5.28 and 2.63, so [Equation (48)](#eq-side-weir-integrated-falling-length) reproduces the bracket 5.28-2.63c/E_w . Both numbers depend on the pressure correction and the assumed end heads; a different h_1/h_2  or h_1/E_w  would change them.

## Match the multiplier 2.03

The remaining multiplier is 1/(k\\sqrt{\\alpha_f}) . With \\alpha_f=1.4 , MSMA’s value 2.03 corresponds to

 k=12.031.4=0.416332,Cd=32k=0.624498k=\\frac1{2.03\\sqrt{1.4}}=0.416332,\\qquad C_d=\\frac32k=0.624498(53)

in our stated strip-law convention. For example, an assumed C_d=0.625  gives 1/(k\\sqrt{1.4})=2.02837 , which rounds to 2.03.

Combining the rounded bracket with that multiplier yields [Equation (54)](#eq-side-weir-reconstructed-msma-falling), the equation printed in MSMA 2:

 L=2.03B(5.28−2.63cEw).L=2.03B\\left(5.28-2.63\\frac{c}{E_w}\\right).(54)

To decide whether the falling-profile formula applies, MSMA 2 uses the criterion c/E_w<0.6 , with E_w  measured above the crest. The entrance and downstream conditions must also allow the water surface to fall.

## MSMA’s pressure correction contradicts its constants

MSMA prints two values for this formula that do not agree with each other:

- The constants 5.28 and 2.63 in the length formula. As shown above, these come out only with a constant pressure correction \\eta=0.8 , the APWA/FWQA value.
- The pressure correction MSMA tells us to use: \\eta=0.95  (MSMA 2 p. 20-18).

If we follow MSMA and use \\eta=0.95  in the same integration, the bracket becomes

 5.10−3.17cEw5.10-3.17\\frac{c}{E_w}(55)

instead of 5.28-2.63c/E_w . MSMA’s printed formula, which we use below, therefore corresponds to \\eta=0.8 .

## Calculate a falling-profile length

[Figure 17](#fig-side-weir-falling-example) shows the approach channel at the top and, below it, the end heads that the falling-profile formula assumes.

Figure 17. Falling example: approach dimensions and the shortcut’s assumed endpoint heads

We now use the falling-profile formula for a second example. Unlike the rising-profile example, we do not specify the end depths: the formula already assumes h_1=E_w/2  and h_2=E_w/20  through the constants just derived. For the approach channel we take B=2.00\\ \\mathrm{m} , y_n=1.00\\ \\mathrm{m} , V_n=1.50\\ \\mathrm{m/s}  and c=0.30\\ \\mathrm{m} . First we calculate E_w  from [Equation (42)](#eq-side-weir-msma-approach-head) with MSMA 2’s approach correction factors:

 Ew=1.21.5022(9.81)+1.0(1.00−0.30)=0.837615 m.E_w=1.2\\frac{1.50^2}{2(9.81)}+1.0(1.00-0.30)=0.837615\\ \\mathrm{m}.(56)

The ratio c/E_w=0.35816  is below MSMA 2’s stated 0.6 criterion. Substitution into [Equation (41)](#eq-side-weir-msma-falling) gives

 L=2.03(2.00)\[5.28−2.63(0.300.837615)\]=17.6124 m.L=2.03(2.00)\\left\[5.28-2.63\\left(\\frac{0.30}{0.837615}\\right)\\right\]=17.6124\\ \\mathrm{m}.(57)

So for this 2.00 m wide channel with a 0.30 m high crest, MSMA’s falling-profile formula gives a crest length of about 17.61 m. Notice that we never specified how much flow to divert; the formula fixes the end heads instead. To find the diversion, we calculate the discharges at both ends of the crest from a falling-profile calculation that agrees with the entrance and downstream conditions, and take their difference, Q_s=Q_1-Q_2 .

If we had mistakenly used the head above the invert, E_w+c=1.137615\\ \\mathrm{m} , the same formula would give about 18.62 m, roughly 1 m too long. E_w  must be measured above the crest.

# Conclusion

In this article, we derived the general crest-length equation for a side-overflow weir from continuity, then applied it to the rising and falling profiles in MSMA. The table below compares each MSMA item with our derivation.

| Water surface | MSMA formula | Our derivation | Agrees with MSMA? |
| --- | --- | --- | --- |
| Rising | MSMA 2 Eq. 20.13: L=(B/C_m)\[\\phi(y_2/E)-\\phi(y_1/E)\] , with C_m=0.35 | Constant specific energy gives L=(B/k)\[\\Psi_2-\\Psi_1\] | Yes. Same formula, with \\phi=\\Psi  and C_m=k . |
| Falling | MSMA 2 Eq. 20.11: L=2.03B(5.28-2.63c/E_w) , with pressure correction \\eta=0.95 | Corrected head with \\alpha_f=1.4 , \\eta=0.8 , h_1=E_w/2 , h_2=E_w/20  and C_d=0.625  gives L=2.03B(5.28-2.63c/E_w) | Yes, but only with a constant pressure correction \\eta=0.8 . MSMA lists 0.95, which would change 5.28 and 2.63. |

Our derivation therefore reproduces MSMA’s formulas, apart from the listed pressure correction of 0.95.

For design, we start with the incoming flow and the amount the downstream drain can carry, then determine the diversion required. The crest height and the changing head along the opening determine how much water each metre can spill, so we integrate to find the required length. For an existing weir, we use its fixed dimensions to calculate the diversion. In either case, the endpoint water levels must agree with the connecting channel and its controls, and the receiving pond must remain low enough for the free-overflow equations to apply.

### *Related*

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I’m the [Benevolent Dictator for Life](https://en.wikipedia.org/wiki/Benevolent_dictator_for_life) for MiTS Software cum Editor of this website. Read more [here](/about-ngu-soon-hui/).

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