On Jul 5, 2012, at 8:34 AM, Boudewijn Verkooijen wrote:
Dear all,
I'm using the curve() function to plot discharge Q against water
depth a.
However, I would like to have a graph of water depth a plotted against
discharge Q. How can this be done?
Minimal working example:
S0 = 0.004
n = 0.04
tanalpha = 1.4/1.5
par(mar = c(5,5,1,1)) # b, l, t, r
curve((sqrt(S0)/n)*(0.035+(0.7+(x-0.1)/
tanalpha)*(x-0.1))*((0.035+(0.7+(x-0.1)/tanalpha)*(x-0.1))/
(2*sqrt((0.7/2)^2+0.1^2)+2*sqrt((x-0.1)^2+((x-0.1)/
tanalpha)^2)))^(2/3),0.1,1.55,
lwd = 3, col = "royalblue4", ann = F, axes = T)
title(xlab = parse(text='a~bgroup("[", m, "]")'))
title(ylab = parse(text='Q~bgroup("[", m^3/s, "]")'))
box()
I tried to find the inverse function, but that doesn't seem to exist.
R does not perform computer algebra. If you wanted a numerical
approach, you can construct a close fit to that function with
approxfun() and then reverse the x and y roles to create an inverse.
Actually, since curve returns a list with x and y components you could
also do this:
xycurv <- curve((sqrt(S0)/n)*(0.035+(0.7+(x-0.1)/
tanalpha)*(x-0.1))*((0.035+(0.7+(x-0.1)/tanalpha)*(x-0.1))/
(2*sqrt((0.7/2)^2+0.1^2)+2*sqrt((x-0.1)^2+((x-0.1)/
tanalpha)^2)))^(2/3),0.1,1.55,
+ lwd = 3, col = "royalblue4", ann = F, axes = T)
plot(xycurv$y, xycurv$x)
--
David Winsemius, MD
West Hartford, CT
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