 So today we're going to talk a little bit about stage-discharge relationships and how we can use data we've gathered or geometric information we've calculated to figure out a relationship between stage and discharge. So Manning's equation and other equations establish a relationship between flow in a channel, which we abbreviate with the letter Q, and it's often called discharge if you think about it exiting a stream bed. Usually we think about this as being at the end of a stream exiting out into some other system. So we often call it discharge. So Manning's equation establishes this relationship between discharge and the geometry of the water in the channel cross-section. So if we have some sort of channel cross-section here and we know what the geometry of the channel is and there's water in it, well we measure some depth of the water or actually more frequently it's often written as the height of the water with above a certain datum. But this is called the stage. The stage of the water depends on how high it is above some sort of datum. And it's effectively the depth of the water at the deepest point in the channel. And if we use Manning's equation, we have a relationship here between Q, which is the discharge or the flow related to the stage or that depth D. And again for example something like Manning's equation 1.49 over N, we have an area which depends upon the depth and we have a hydraulic radius which depends upon the depth. We take that to the two-thirds power and then we also have a slope. So there again there are different relationships and if we use Manning's relationship, if we have a given channel location with a known geometry then we can actually simplify this equation into parts that relate to the depth D and to parts that are constant. For example for a given channel my Manning's coefficient will be a constant and my slope upstream of the channel will be a constant. And then if we look at the cross section for example let's say it is a triangular cross section like I've drawn here and perhaps we can parameterize it with some constant values. For example an angle here, angle theta, then our relationships such as our area, our area here might be something like the depth square divided by the tangent of the angle. If we're looking for the area of a triangle and again labeling this as being our depth. Okay the cross sectional area looks like that. The wetted perimeter can be found using a relationship two times the depth divided by the sign of that angle. And finally the hydraulic radius which is the ratio of those two things can be calculated to be the depth times the cosine of the angle divided by two. So we can replace this area here in Manning's equation and the hydraulic radius and we end up with an equation for Q as a function of the depth D that basically 1.49 over N square root of S naught the cosine of the angle to the two-thirds power and then two to the two-thirds power times the tangent of the angle. And all I've done here is I've gathered all the constant pieces, all the constant pieces and put them into sort of one big number and then there's some dependency and if you do the algebra here you'll find that it's D to the eight-thirds power. So now we have a dependency upon the stage or the depth of the stage and then we have some large constant value that takes into account all the other measurements in the geometry. And notice this would change if we had a different shape of the channel cross-section. But the idea here is that there is a relationship between the stage and the discharge. Now more generally we can write it this way Q the discharge is equal to some coefficient times the stage to the power of some exponent. In this case you could see where the exponent would be eight-thirds and the coefficient would be dependent upon all those other values. There's our coefficient and here we have an exponent. And notice we could plot that relationship by plotting the stage on the x-core axis and plotting the flow of the discharge and the y-axis. And we would get some sort of version of something that looks like a parabolic equation but basically depending on the steepness of that slope would depend on that exponent. So the interesting thing about equations of this form in this sort of exponential form is that we can manipulate them using logarithms. And this is a pretty common tactic when we're trying to analyze complex systems. I'm going to go ahead and take the logarithm of both sides of this equation. And when I do so I get the logarithm of Q is equal to the logarithm of the other side AD to the B power. Notice I could take logarithm base 10 or natural logarithms or actually any logarithm but in this case we're going to go ahead and assume logarithm base 10. And if we use some logarithmic rules we know for example that the logarithm of two things that are multiplied is equal to the sum of the two logarithms, logarithm of x plus logarithm of y. That's one of the logarithm rules we use. And if we put that in here we can rewrite this equation log Q is equal to the log of D to the B plus the log of A. Notice I've swapped the A and the D to the B in this case which will be evident y in just a moment. And another logarithmic rule that we can take advantage of is recognizing that the logarithm of x to the power of k is the same as k times the logarithm of x. In other words you can move this down in front, the exponent in front of the logarithm. It's the same as multiplying. And if we do that process we can now rewrite this equation again. The logarithm of Q is equal to B logarithm of D plus logarithm of A. Now here's why these substitutions were valuable. What we've done is we've taken a somewhat more complicated equation that had an exponent and a coefficient in it and reduced it to being a linear equation. Notice this is in the form, if we sort of ignore the logs for the moment, this is in the form y equals m x plus B as long as we recognize that the logarithm of Q could be considered as our y value. The logarithm of D is our x value, the logarithm of A is our B value, and the slope is that large B that we have there. The fact that this is a linear equation is particularly useful because if we have a series of discharge values, Q, and we have a list of some discharge values here, and we have a series of values D that correspond with those depths and discharges, what we can find is a relationship between these two things. But what we're going to do instead of finding the relationship that we have here, where we have some curve that we don't know about, what we're going to choose to do instead is we're going to plot the logarithm of D on my x-axis and the logarithm of Q on the y-axis. When we do so, if our data really fits this exponential form, we will get something that looks linear on the logarithmic curve. And then what we can choose to do with that curve is to fit the data with a line in the form y equals mx plus b, noting that what was on the x-axis was indeed our logarithm of D and what was on the y-axis was our logarithm of Q. And then whatever slope we end up measuring there, we know that that becomes our exponent b, and whatever intercept, we find the slope, that's our exponent b, and then whatever intercept we are able to find from fitting the equation, we know is equal to the logarithm of our value a. If I plug those things back into sort of my initial equation up here, Q equals a D over b, let me rewrite that, Q equals a D over b, let me sort of plug those pieces back in, and instead I recognize that Q is equal to D to the m, in other words the slope of this line. And then I recognize that if logarithm of a is equal to b, I'm going to use the inverse of both of those, which is basically taking them, 10 to the exponent is the inverse function of the logarithm, and I can write it this way, 10 to the b power. So if I can find the m and I can find the b, I can rewrite my equation as Q is equal to 10 to the b power times D to the mth power.