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.
Thank you for support,
Boudewijn
[[alternative HTML version deleted]]
On 05/07/2012 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?curve() is designed for plotting y vs x where y is a function of x, so I think you need to do it yourself. Simply compute a vector of "a" values using seq(), compute the corresponding Q values using your formula, and then use plot(Q, a, type="l") You can add the lwd and other arguments too if you like. Duncan Murdoch> 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. > Thank you for support, > Boudewijn > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
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