Can you please explain cut off ratio as described in Heat and Thermodynamics by Mark W Zemansky and Richard H Dittman 
This is what said in Heat and Thermodynamics written by Richard H Dittman and Msrk W Zemansky about cut off ratio of Diesel engine. According to Zemansky cut off ratio is $\frac{V_1}{V_2}$ But when I searched web it is found to be $\frac{V_3}{V_1}$.
Is there any way I can relate both. 
Since this is a standard textbook for many universities and a question raised many of my colleagues, I expect someone would answer this question. 
 A: It is a misprint in Dittman and Zeamnsky's book. Even the best books, however often one reads and rereads the proofs, still contain misprints.
Comparing (6.6) with the first definition of $\eta$ one sees that 
$r_E=T_3/T_2$ from the denominator and
$r_E^\gamma=T_4/T_1$ from the numerator.
Consider the first relation: the evolution from point 2 to point 3 is at constant pressure. Therefore the volumes are proportional to the temperatures, so 
$r_E=T_3/T_2=V_3/V_2$
is indeed correct.
Now imagine the left top part of the picture fixed. The points 2 and 3 are fixed, and thus $r_E$ is fixed. What prevents me from making the two adiabatics longer, or shorter, stopping at whatever value of $V_1$ that I choose (larger than $V_3$) ? Since the complicated factor in $\eta$ does not depend on $r_C=V_1/V_2$, there is no contradiction. Whatever value of $V_1$ I end up with, this factor in the efficiency is the same.  Of course this will change $P_0$ but $P_0$ does not appear anywhere !
The second relation, the one from the numerator, involves $T_1$ that does depend on $V_1$ but only through the ratio $T_4/T_1$. It is easy to prove that though both $T_1$ and $T_4$ depend on $V_1$, their ratio does not. It depends only on $V_3/V_2$. 
Here is the proof. To make things clearer, I use $P_1=P_0$.
The equations of the adiabatics are 
$P_4 V_4^\gamma=P_3 V_3^\gamma$
$P_1 V_1^\gamma=P_2 V_2^\gamma$
But $V_1=V_4$, and $P_2=P_3$ so 
$P_4 V_1^\gamma=P_2 V_3^\gamma$
$P_1 V_1^\gamma=P_2 V_2^\gamma$
Take the ratios, you find $P_4/P_1=(V_3/V_2)^\gamma$
Now the points 1 and 4 are at the same volume. Hence the ratio of the pressures is equal to the ratio of the temperatures. Hence 
$T_4/T_1=P_4/P_1=(V_3/V_2)^\gamma=r_E^\gamma$
as expected. $V_1$ does not appear in the ratio $T_4/T_1$ even though of course both $T_1$ and $T_4$ depend on $V_1$.
$V_1$ can be choosen completely arbitrarily, provided it is larger than $V_3$ of course.
Misprints survive in published work, you can take my word for it. I have plenty of misprints in my published articles, proofreading is never perfect.
Congratulations for finding this one !
Ooops ! There is (at least) one misprint in this answer. I did not do it on purpose, it is perfectly a genuine "honest mistake". I reread this post many times and I missed it till now. Since it does not at all affect the reasoning, I'm leaving it for you to find....
EDIT : 
Well, there were (at least) two misprints. The most severe was in fact more than a misprint.  The one I found last I corrected in italics.
I had erroneously called $\eta$ the complicated factor in $\eta$ that depends on $r_E$ but indeed  not in $r_C$ or on $V_1$. Of course, $\eta$ itself does depend on $V_1$ (or $r_C$) but only through the simple factor $T_1/T_2$. Indeed on the adiabatic curve one has 
$P_1 V_1^\gamma=P_2 V_2^\gamma$
and thus 
$T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}$
so $T_1/T_2=(V_2/V_1)^{\gamma-1}=r_C^{1-\gamma}$
But my reasoning is still correct, I just concentrated on the complicated factor and mistakenly called it $\eta$ when it was just one factor in $\eta$. Sorry about that. On the other hand, the same lack of precision is already in the book itself ! It says there that "the efficiency (...) does not depend on $r_C$", and this must be understood as "the complicated factor". The ratio $T_1/T_2$ in the efficiency is not independent on $r_C$ !
The other misprint, the one I saw the previous time, I left on purpose. That one really has nothing at all to do with the physics involved.
