---
title: "MOC numerical, characteristics not on diagonal (30 cm)"
subject: "Computational Techniques in Civil Engineering"
chapter: 6
key: "semester_8/computational_techniques_in_civil_engineering:moc-numerical-off-diagonal-a2"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/moc-numerical-off-diagonal-a2
site: BCT Engineering
---

# MOC numerical, characteristics not on diagonal (30 cm)

Computational Techniques in Civil Engineering · Chapter 6: Method of Characteristics

- Asked in: 2072 Asoj, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/moc-numerical-off-diagonal-a2
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-6

## Question

If the MOC is applied for $t_1 = 1$ sec and $t_2 = 2$ sec, time levels for a pipe with diameter 30 cm carrying water. If $Q_A = 0.7$ m$^3$/s, $Q_B = 0.76$ m$^3$/s and $Q_C = 0.74$ m$^3$/s, $H_A = 20$ m, $H_B = 20.6$ m and $H_C = 20.4$ m are the values at grid points. Find the values of $Q$ and $H$ at $t_1 = 1$ sec that will be required for finding $Q$ and $H$ at P when characteristics do not lie on diagonal. Here, $\Delta x = 1000$ m, $\Delta t = 1$ sec, $f = 0.02$ and $c = 800$ m/s.

## Short answer

theta=a dt/dx=0.8. Interpolate R between A and C: Q_R=0.708 m^3/s, H_R=20.08 m; S between C and B: Q_S=0.756 m^3/s, H_S=20.56 m. Use these in the C+ and C- equations with friction length a dt=800 m.

## Answer

#### Idea
When the characteristics do not lie on the diagonal of the grid, $a\Delta t<\Delta x$: the $C^+$ and $C^-$ lines through P (at $t_2$) meet the line $t_1$ at points R and S that lie **between** the grid points, not at A and B. $Q$ and $H$ at R and S are found by linear interpolation along the time line $t_1$.

```text
 t2 :              P
                  /|\
            C+  /  |  \  C-
 t1 :   A-----R----C----S-----B
        (x-dx)         (x+dx)
        |<-a dt->|   |<-a dt->|
```
Here R lies on the $C^+$ line between A and C, and S on the $C^-$ line between C and B, each at a distance $a\Delta t$ from C.

#### Interpolation factor
$$
\theta=\frac{a\,\Delta t}{\Delta x}=\frac{800\times1}{1000}=0.80
$$
**Point R (between A and C, $C^+$ foot):**
$$
Q_R=Q_C-\theta\,(Q_C-Q_A),\qquad H_R=H_C-\theta\,(H_C-H_A)
$$
$$
Q_R=0.74-0.80(0.74-0.7)=0.708\ \text{m}^3/\text{s},\qquad
H_R=20.4-0.80(20.4-20)=20.08\ \text{m}
$$
**Point S (between C and B, $C^-$ foot):**
$$
Q_S=Q_C-\theta\,(Q_C-Q_B),\qquad H_S=H_C-\theta\,(H_C-H_B)
$$
$$
Q_S=0.74-0.80(0.74-0.76)=0.756\ \text{m}^3/\text{s},\qquad
H_S=20.4-0.80(20.4-20.6)=20.56\ \text{m}
$$

**Answer:** at $t_1=1$ s, $Q_R=0.708$ m$^3$/s, $H_R=20.08$ m on the $C^+$ line, and $Q_S=0.756$ m$^3$/s, $H_S=20.56$ m on the $C^-$ line. These replace $Q_A,H_A$ and $Q_B,H_B$ in the $C^+$ and $C^-$ equations
$$
H_P=H_R-B_p(Q_P-Q_R)-R_pQ_R|Q_R|,\qquad H_P=H_S+B_p(Q_P-Q_S)+R_pQ_S|Q_S|
$$
with $B_p=a/(gA_p)$ and $R_p=f\,a\Delta t/(2gDA_p^2)$ (the friction length is $a\Delta t=800$ m).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why must Q and H at points R and S be found by interpolation?
   A. Because a dt < dx, so the characteristic feet fall between grid points
   B. Because friction is large
   C. Because the pipe diameter changes
   D. Because a dt > dx, so feet fall outside the grid
   Answer: A. When a dt < dx the C+ and C- lines meet time t1 between the nodes, so values are interpolated. (Key point: Off-diagonal characteristics)

2. What is theta = a dt/dx here and what does it represent?
   A. 1.25, ratio of dx to a dt
   B. 0.80, the fraction of the grid interval the foot lies from C
   C. 800, the friction length
   D. 0.0125, the ratio dt/a
   Answer: B. theta = 800 x 1/1000 = 0.80, the fraction of dx from C to the foot. (Key point: Interpolation factor)

3. What is Q_R on the C+ line using Q_R = Q_C - theta(Q_C - Q_A)?
   A. 0.756 m^3/s
   B. 0.700 m^3/s
   C. 0.732 m^3/s
   D. 0.708 m^3/s
   Answer: D. Q_R = 0.74 - 0.8(0.74 - 0.70) = 0.708 m^3/s. (Key point: Value at R)

## Close variants set in other exams

- [MOC numerical, characteristics not on diagonal (25 cm)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Amoc-numerical-off-diagonal-j) — Computational Techniques in Civil Engineering, chapter 6, asked 2×

## Related in Computational Techniques in Civil Engineering

- [MOC numerical: discharge and head at P (30 cm pipe)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Amoc-pipe-discharge-head-pc-30cm) — Computational Techniques in Civil Engineering, chapter 6