R help - Mar 2010 - getting CI's for certain y of nls fitted curve

hello,

i managed to get CI's for my curve - but now I need the intervall for a
certain y point (y_tenth) of the curve..
can anyone help me with this?

#####data:
por<-data.frame(list(structure(list(run = structure(c(1L, 1L, 1L, 1L, 2L,
2L,
2L, 2L, 3L, 3L, 3L, 3L), .Label = c("1", "3",
"4"), class = "factor"),
    press = c(15, 21, 24, 29.5, 15, 21, 24, 29.5, 15, 21, 24,
    29.5), tr_rel = c(1, 0.459454191, 0.234697856, 0.135282651,
    1, 0.853283066, 0.306314797, 0.186302231, 1, 0.42980063,
    0.103882476, 0.086463799), tr = c(513, 235.7, 120.4, 69.4,
    318.3, 271.6, 97.5, 59.3, 476.5, 204.8, 49.5, 41.2)), .Names =
c("run",
"press", "tr_rel", "tr"), row.names = c(NA, -12L),
class = "data.frame")))

summary(mod1<-nls(tr ~ SSlogis( log(press), Asym, xmid, scal),data=por))

press_x <- seq(10, 40, length = 100)

predict(mod1, data.frame(press = press_x))

with(por, plot(press,tr,xlim=c(10,35),ylim=c(0,500)))

lines(press_x, predict(mod1, data.frame(press = press_x)))

###http://finzi.psych.upenn.edu/R/Rhelp02/archive/42932.html
###calc. of conf. intervalls, by linear approximation
se.fit <- sqrt(apply(attr(predict(mod1, data.frame(press
press_x)),"gradient"),1,
function(x) sum(vcov(mod1)*outer(x,x))))

###plot points and curve with ci's
matplot(press_x, predict(mod1, data.frame(press
press_x))+outer(se.fit,qnorm(c(.5, .025,.975))),
type="l",lty=c(1,3,3),ylab="transp",xlab="waterpot")
with(por, matpoints(press,tr,pch=1))

###the desired y is the tenth of the y asymptote

asym<-coef(mod1)[1] 

tenth<-asym*0.1
abline(h=asym)
abline(h=tenth,lty=5)
text(10,65,"10% of max. transpiration",adj=0)

###finding corresponding x by y with inverse function:
Asym=coef(mod1)[1]
xmid=coef(mod1)[2]
scal=coef(mod1)[3]

y_tenth<-asym*0.1
x_tenth<-exp((xmid-log(Asym/y_tenth-1)*scal));names(x_tenth)<-"x"
x_tenth

abline(v=x_tenth)
text(10,25,paste("at waterpot.:",
round(x_tenth,2)),adj=0)

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...it's of course simply using the desired x in the predict function. 
in this case: predict(mod1,data.frame(press = x_tenth[1]).

it must have been a trivial syntax error, why this didn't work in the first
place.

kay
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