This method is generally considered "old-fashioned" so while it may be possible, the syntax is difficult and I suspect fewer people know how to manipulate the anova commands to get what you want. The more common method is using glht with a likelihood-based model from nlme or lme4. (I'm certainly welcome to be proved wrong by other answers though.)
That said, if I needed to do this, I wouldn't bother with the anova commands; I'd just fit the equivalent model using lm, pick out the right error term for this contrast, and compute the F test myself (or equivalently, t test since there's only 1 df). This requires everything to be balanced and have sphericity, but if you don't have that, you should probably be using a likelihood-based model anyway. You might be able to somewhat correct for non-sphericity using the Greenhouse-Geiser or Huynh-Feldt corrections which (I believe) use the same F statistic but modify the df of the error term.
If you really want to use car, you might find the heplot vignettes helpful; they describe how the matrices in the car package are defined.
Using caracal's method (for the contrasts 1&2 - 3 and 1&2 - 4&5), I get
psiHat tStat F pVal
1 -3.0208333 -7.2204644 52.1351067 2.202677e-09
2 -0.2083333 -0.6098777 0.3719508 5.445988e-01
This is how I'd get those same p-values:
Reshape the data into long format and run lm to get all the SS terms.
library(reshape2)
d <- OBrienKaiser
d$id <- factor(1:nrow(d))
dd <- melt(d, id.vars=c(18,1:2), measure.vars=3:17)
dd$hour <- factor(as.numeric(gsub("[a-z.]*","",dd$variable)))
dd$phase <- factor(gsub("[0-9.]*","", dd$variable),
levels=c("pre","post","fup"))
m <- lm(value ~ treatment*hour*phase + treatment*hour*phase*id, data=dd)
anova(m)
Make an alternate contrast matrix for the hour term.
foo <- matrix(0, nrow=nrow(dd), ncol=4)
foo[dd$hour %in% c(1,2) ,1] <- 0.5
foo[dd$hour %in% c(3) ,1] <- -1
foo[dd$hour %in% c(1,2) ,2] <- 0.5
foo[dd$hour %in% c(4,5) ,2] <- -0.5
foo[dd$hour %in% 1 ,3] <- 1
foo[dd$hour %in% 2 ,3] <- 0
foo[dd$hour %in% 4 ,4] <- 1
foo[dd$hour %in% 5 ,4] <- 0
Check that my contrasts give the same SS as the default contrasts (and the same as from the full model).
anova(lm(value ~ hour, data=dd))
anova(lm(value ~ foo, data=dd))
Get the SS and df for just the two contrasts I want.
anova(lm(value ~ foo[,1], data=dd))
anova(lm(value ~ foo[,2], data=dd))
Get the p-values.
> F <- 73.003/(72.81/52)
> pf(F, 1, 52, lower=FALSE)
[1] 2.201148e-09
> F <- .5208/(72.81/52)
> pf(F, 1, 52, lower=FALSE)
[1] 0.5445999
Optionally adjust for sphericity.
pf(F, 1*.48867, 52*.48867, lower=FALSE)
pf(F, 1*.57413, 52*.57413, lower=FALSE)
Best Answer
Functionally, they operate similarly.
From this page (demo in SPSS)
http://www.uvm.edu/%7Edhowell/StatPages/More_Stuff/RepMeasMultComp/RepMeasMultComp.html