diff --git a/docs/Proportional-Interval-Stamp/Proportional-Interval-Stamp.mdx b/docs/Proportional-Interval-Stamp/Proportional-Interval-Stamp.mdx
index 2b0e7a6..0de4768 100644
--- a/docs/Proportional-Interval-Stamp/Proportional-Interval-Stamp.mdx
+++ b/docs/Proportional-Interval-Stamp/Proportional-Interval-Stamp.mdx
@@ -175,28 +175,28 @@ and we already know the radius at a segment $r(x) = r_0 - \cos\theta x$.
 
 We need the stamp interval to be proportional to radius.
 Therefore, for each segment, define the stamp interval as $\eta r(x)$.
-To find the number of stamps $\Delta N$ in a segment, divide the length by the interval:
+To find the number of stamps $\Delta n$ in a segment, divide the length by the interval:
 
 $$
 \Delta n = \frac{\Delta x}{\eta r(x)}
 $$
 
-To find the number of stamps $\Delta n$ over the edge, we need to integrate $\Delta n$ from 0 to x.
+To find the number of stamps $\Delta n$ over the edge, integrate $\Delta n$ from 0 to x.
 
 $$
 n(x) = \int_0^{x} \frac{dx}{\eta r(x)}
 $$
 
-Given $r(x) = r_0 - \cos\theta x$, we substitute this into the integral and get:
+Given $r(x) = r_0 - \cos\theta x$, substitute it into the integral and get:
 
 $$
-\tag{1} n(x) = \frac{1}{\eta \cos\theta} \ln(r_0 - \cos\theta x) \bigg|_0^x = \frac{1}{\eta \cos\theta} \ln (1 - \frac{\cos\theta x}{r_0})
+\tag{1} n(x) = -\frac{1}{\eta \cos\theta} \ln(r_0 - \cos\theta x) \bigg|_0^x = -\frac{1}{\eta \cos\theta} \ln (1 - \frac{\cos\theta x}{r_0})
 $$
 
 Exchange the dependent and independent variable in formula (1) and get:
 
 $$
-\tag{2} x(n) = \frac{r_0}{\cos\theta}(1 - e^{\eta\cos\theta n})
+\tag{2} x(n) = \frac{r_0}{\cos\theta}(1 - e^{-\eta\cos\theta n})
 $$
 
 As $x = L$, we know the total stamp number on the edge, remind that $\cos\theta L = r_0-r_1$:
@@ -205,21 +205,67 @@ $$
 \tag{3} n(L) = \frac{1}{\eta \cos\theta}\ln \frac{r_0}{r_1}
 $$
 
-Soon we will use the formula (1)(2)(3) in our code.
+We will soon use the formula (1)(2)(3) in our code.
 
-![store](store-n.png)
-
-How do we place footprint on the polyline?
+How do we place footprints?
 Imagine a point starts from the first vertex and moves along the polyline.
-The stamp number to the first vertex grows with its movement.
-Every time the number hits an integer, we place a footprint at this point.
+Its stamp index (number of stamps from the first vertex) $n$ grows with its movement.
+Each time this index hits an integer, place a footprint at that point.
 
-Remind the figure below in the Stamp section.
-Given a point or pixel, we calculate a range on the edge that can cover the point.
-We label the two roots with $x_1$ and $x_2$.
+Remind the below figure in the Stamp section.
+Given a point, we calculate a range on the edge that can cover the point by solving a quadratic equation.
+Label the two roots with $x_1$ and $x_2$.
 
 ![locate stamp](./locate-stamp.png)
 
-With the stamp number of each edge $n(L)$, we can prefix sum $n(L)$ the number of stamps from each vertex to the first vertex.
-We will store the number into each vertex.
-For the current edge, we use $n_0$ and $n_1$ to indicate the stamp number at vertex0 and vertex1.
+To calculate the nearest stamp point (the black dot in the figure) between $x_1$ and $x_2$,
+we need to compute the number of stamps from $x_1$ to the polyline's first vertex, which is also called stamp index at $x_1$.
+According to formula (1), on the current edge, the number of stamps from vertex0 to $x_1$ is $n(x_1)$.
+Label the stamp index of vertex0 as $n_0$, and label vertex1's as $n_1$.
+
+![store](./store-n.png)
+
+Therefore, the stamp index of $x_1$ is $n_0 + n(x_1)$.
+Because we place footprints at the points with integer stamp index,
+the ceiling of $n_0 + n(x_1)$ is the nearest stamp's index,
+denoted as $n_\mathrm{nearest} = \lceil n_0 + n(x_1) \rceil$.
+Replace it into the formula (2) to get its position $x_{\mathrm{nearest}} = x(\lceil n_0 + n(x_1) \rceil)$,
+and all other positions $x(n_\mathrm{nearest} + 1)$, $x(n_\mathrm{nearest} + 2) \dots$
+
+Additionally, to determine $n_0$ and $n_1$, we can compute the prefix sum of the $n(L)$ in formula (3) over all edges.
+Put them into vertex data and pass them into fragment shader, exactly same as the value `length` in the Stamp section.
+
+## Implementation
+
+## Corner case
+
+## Proof of properties
+In the Stamp Pattern subsection, we claim that "stamp interval is always proportional to radius of stroke" and
+"proportional interval has more consistent appearance".
+Though they are intuitively correct, it's time to prove them mathematically!
+
+### Proportional interval
+**Conclusion:**
+Pick two consecutive footprints on an edge.
+Label their stamp indices as $n$ and $n+1$, positions as $x(n)$ and $x(n+1)$, radii as $r(n)$ and $r(n+1)$.
+Their interval $x(n+1) - x(n)$ is proportional to both $r(n)$ and $r(n+1)$:
+$$
+\begin{aligned}
+\frac{x(n+1) - x(n)}{r(n)} &= \frac{1 - e^{-\eta\cos\theta}}{\cos\theta}\\
+\frac{x(n+1) - x(n)}{r(n+1)} &= \frac{e^{\eta\cos\theta} - 1}{\cos\theta}
+\end{aligned}
+$$
+
+**Proof:**
+Take the formula (3) into $r(x) = r_0 - \cos\theta x$ to get $r(n) = r_0e^{-\eta\cos\theta n}$.
+The rest of proof is straightforward.
+
+Notice the ratios do not depend on variable $n$.
+I was amazed by this result.
+Even now, it still fascinates me that how a naive rule (interval is proportional to the radius continuously) can produce an elegant result.
+
+### Consistent appearance
+**Conclusion:**
+For points that have the same stamp number that can cover it (i.e., $n(x_2) - n(x_1)$ value is the same),
+they form a straight line.
+I call this line "isocolor" line (analogy to "isoheight" line).
\ No newline at end of file