Obstacle in chaining may be classified as

1) When chaining is free but vision is obstructed

2) When chaining is obstructed but vision is free

3) When chaining and vision are both obstructed

1) When chaining is free but vision is obstructed -

Such a problem arise when a rising ground or a jungle area interrupt the chain line

The end stations are not visible from intermediate points when a jungle area come across the chain line.

Let $AB$ is the actual chain line which cannot be ranged and extended because of interruption of jungle. Let the chain line extended up to $R$. a point $P$ select on the chain line and random line $PT$ is taken in a suitable direction. Point $C$, $D$ and $E$ are selected on the random line and perpendiculars are projected from them. Perpendicular point $C$ meet the chain line at $C_1$. The perpendicular point $D$ and $E$ meet the chain line $D_1$ and $E_1$. Now the distance $PC$, $PD$, $PE$ and $CC_1$ are measured From triangle $PDD_1$ and $PCC_1$ $$ \frac{DD_1}{PD}=\frac{CC_1}{PC} \ DD_1=\frac{CC_1}{PC}\times{PD}$$

Again, from triangle $PEE_1$ and $PCC_1$

$$ \frac{EE_1}{PE}=\frac{CC_1}{PC} \\ EE_1=\frac{CC_1}{PC}\times{PE}$$

Length $DD_1$ and $EE_1$ is calculated. The calculated distance measured along the perpendicular at $D$ and $E$ Point $D_1$ and $E_1$ should lie in the chain line $AB$, which can be extended accordingly. $$distance\ PE_1=\sqrt{{PE}^2+{EE_1}^2 }$$

2) When chaining is obstructed but vision is free Such a problem arise when a pond or a river come across the chain line

Suppose $AB$ is the chain line .two point $C$ and $D$ are selected on it on opposite bank of the pond. Equal perpendicular $CE$ and $DF$ are erected at $C$ and $D$. the distance $EF$ measured. $$CD=EF$$

The pond may also be crossed by forming a triangle as shown in fig. a point $C$ is select on a chain line. The perpendicular $CE$ set out at $C$ and line $ED$ is suitably taken. The distance $CE$ and $ED$ are measured. $$CD=\sqrt{{ED}^2-{CE}^2}$$

A small river comes across the chain line. Suppose $AB$ is chain line. Two point $C$ and $D$ are selected on this line on opposite bank of the river. At $C$ a perpendicular $CE$ is erected and bisected at $F$. a perpendicular is set out at $E$ and a point $G$ is selected on it that $D$, $F$ and $G$ are in the same straight line. From triangle $DCF$ and $GEF$ $$GE=CD$$

The distance $GE$ is measured and thus the distance $CD$ is obtained indirectly.

3) When chaining and vision are both obstructed Such a problem arise when a building come across the chain line. Suppose $AB$ is the chain line. Two point $C$ and $D$ are selected on it at one side of the building. Equal perpendicular $CC_1$ and $DD_1$ are erected. The line $C_1D_1$ is extended until the building is crossed. On the extended line two poin $E1$ and $F1$ are selected. The perpendicular $E_1E$ and F1F are so erected that $$E_1E = F_1F= D_1D=C_1C $$

Thus the point $C$, $D$, $E$ and $F$ will lie on the same straight line $AB$ Here $$DE=D_1E_1$$ The distance $D_1E_1$ is measured and is equal to the required distance $DE $