Why the apparent length looks greater in the moving frame? (Taylor's Classical Mechanics example 15.3) I was reading a special relativity section in this textbook and I am confused about this part in relativistic snake example (15.3).

The problem statement said the position of the right end of the cleaver, viewed from the student's rest frame, $x_R$ is 100cm. But if this position is viewed from the snake's moving frame, the textbook said it should be $x_R'=125cm$. It can't be right to be because my intuition tells me that $x_R'<x_R$.
I think there is an error in Lorentz transformation calculation.
$$x_R'=\gamma (x_R - Vt_R)$$
$x_R'=125cm$ is derived with values $x_R=100cm$ and $t_R=0$. I think the $t_R=0$ part is wrong because he should use $t_R'=0$ which will give the correct result for $x_R'$: 80cm.
Or maybe I am missing something?
Edit: Here's my spacetime diagram

The length of the green line is 80cm and the total length of the green line and the blue line is 100cm.
The length of the yellow line is 100cm. That means the total length of the yellow line and the purple line is 125cm?
No wait, I think the pink line corresponds to 125cm?
 A: It's good that you drew a spacetime diagram.

Here's a spacetime diagram (which agrees with Taylor's calculation).
It's drawn on "rotated graph paper" so that we can more easily visualize the ticks along the segments.
Each diamond represents 20cm in spatial width.
In the inertial rest frames,

*

*The proper-length of the snake is $OX'=100\rm cm$ (5 diamonds),
whose tail and head travel at $v=(3/5)c$.

*The left and right cleavers are separated by $100\rm cm$ (5 diamonds) in the student(LAB)-frame,
whose worldlines are vertical in this diagram.

Now, each observes the other:

*

*The student measures the apparent length of the snake (the GREEN-segment separation between the red worldlines according to the student) to be $80\rm cm$ (4 diamonds),
in accordance with length contraction.

*The snake measures apparent separation between the vertical cleaver-worldlines (the YELLOW-segment) to be $80\rm cm$ (4 diamonds), in accordance with length contraction.

Note that the GREEN and YELLOW segments have equal Minkowski-lengths,
although they have unequal Euclidean-lengths on the student's spacetime diagram.
The curve of equal separation from the origin event are Hyperbolas, not circles.

The event $S$ is the event that is, according to the snake,
in the same location as the "Right-Cleaver falls"-event $R_C$.
$R_C S$ is parallel to the snake worldline.
It is located $6.25\rm\  diamonds= 125\rm\ cm$, which can be seen by counting.
One can reason trigonometrically.

*

*From Minkowski-right-triangle $L_C TQ$, the snake moves with velocity $(v/c)=\tanh\phi=\frac{OPP}{ADJ}=\frac{TQ}{L_C T}=\frac{3}{5}$.


*The time-dilation factor $\gamma=\cosh\phi=\frac{ADJ}{HYP}=\frac{L_C T}{L_C Q}=\frac{5}{4}$.


*$L_C X'=5\rm\ diamonds=100\rm cm$ is the proper-length of the snake.


*From Minkowski-right-triangle $L_C X' H$ (which is similar to $L_CTQ$), where $L_C X'$ is the adjacent side,
the hypotenuse (opposite the right-angle at $X'$) is the GREEN segment
$$L_C H=HYP=ADJ\frac{1}{\cosh\phi}=L_CX' \frac{1}{\gamma}= {5}\frac{1}{(5/4)}=4\rm\ diamonds=80\rm cm$$.


*From Minkowski-right-triangle $L_C R_C W$ (which is similar to $L_CTQ$), where $L_C R_C$ is the adjacent side,
the hypotenuse (opposite the right-angle at $R_C$) is the YELLOW segment
$$L_C W=HYP=ADJ\frac{1}{\cosh\phi}=L_CR_C \frac{1}{\gamma}= {5}\frac{1}{(5/4)}=4\rm\ diamonds=80\rm cm$$.


*Note that $L_C X'H$ and $L_C S R_C$ are similiar Minkowski-right-triangles.
So, $L_C S$ can be computed by proportionality.

Your worldline of the Right-Cleaver (the ORANGE worldline) doesn't look correct to me---it should meet the YELLOW segment.
Your worldline of the Head-of-the-Snake (the front RED worldline) doesn't look correct to me---it should meet the ORANGE segment earlier.

*

*Right now, I'm not sure how to quickly resolve it.

*That intersection event is spacelike-related to $L_C$.

Imagine an intermediate frame of reference, which views the student moving to the left and the snake moving to the right.
That intersection-event is simultaneous with $L_C$ in that intermediate frame.

Maybe you can share your reasoning on HOW you drew your spacetime diagram.
