The Horizontal and Vertical detailed calculation displayed is based on the road design below.

Road Textual Report Sample here.

** Note: The detailed textual report is available from MiTS 3.3 and above.

Components of Horizontal Alignment #

It is the road’s drawn path on a flat map, consisting of straight lines connected by curves. The horizontal alignment must be designed correctly for the safety and comfort of drivers and passengers.

Details of the curve:

Curve Type: Spiral

Curve: IP2

Central Angle of Circular Curve: 0°56’ 32”

Radius (R_c): 30 m

Design Speed: 30 kph

Super Elevation (SE): 6.00%

Degree of Curvature #

D_{c}=\frac{100}{R_{c}}

Where:

{R_{c}} = Radius of circular curve, in radians

For example,

$$
\begin{aligned}
\displaystyle D_c &= \frac{100}{0.523599} = 190.9858499 \\
\displaystyle D_c &= 190^\circ 59’9″
\end{aligned}
$$

Spiral length from TS to SC or ST to CS, Ls #

For a spiral curve, Spiral Length is equivalent to Runoff Length (the value obtained depends on the calculation type here)

For example, the calculation type is Fixed Mode with Adjustment Factor

$$
\begin{aligned}
L_s &= \text{Input Spiral Length} \times \text{Adjustment Factors} \\
L_s &= 20 \times 1.5 = 30.000\,\text{m}
\end{aligned}
$$

Central angle of Spiral arc Ls (Spiral angle), \theta _{s} #

\theta _{s}=\frac{L_{s}\times D_{c}}{200}

Where:

L_{s} = Spiral length from TS to SC or ST to CS

D_{c} = Degree of circular curve

For example,

$$
\begin{aligned}
\theta _{s} &=\frac{30.000\times 190.9858499}{200}=28\circ 38’52.4″
\end{aligned}
$$

Length of Circular curve, L_{c} #

L_{c}=R_{c}\times\Delta _{c}

Where:

R_{c} = Radius of circular curve

\Delta _{c} = Central angle for circular curve; in radians

For example,

$$
\begin{aligned}
\Delta_{c} &= \Delta – 2\theta_{s} = 111.297699 \times (2 – 28.64787749) = 54.00194402 = 54^{\circ} 0′ 7”
\end{aligned}
$$

$$
\begin{aligned}
L_{c} &= 30 \times 0.942817 = 28.285 \text{ m}\end{aligned}
$$

Tangent distance along initial tangent of any point on Spiral with reference to TS or ST, X_{s} #

X_{s}=L\left(1-\frac{\theta^{2}}{10}+\frac{\theta^{4}}{216}-\frac{\theta^{6}}{9360}+\frac{\theta^{8}}{685440}\right)

Where:

L = Spiral Length from TS to SC or ST to CS

\theta = Central angle of spiral arc, L_{s} (Spiral angle); in radians

For example,

$$
\begin{aligned}
X_{s} &=30.000\left(1-\frac{0.49993^{2}}{10}+\frac{0.49993^{4}}{216}-\frac{0.49993^{6}}{9360}+\frac{0.49993^{8}}{685440}\right) \\
X_{s} &=29.259\,\text{m}
\end{aligned}
$$

Tangent offset from initial tangent of any point on Spiral, Y_{s} #

Y_{s}=30.000\left(\frac{\theta}{3}-\frac{\theta^{3}}{42}+\frac{\theta^{5}}{1320}-\frac{\theta^{7}}{75600}+\frac{\theta^{9}}{6894720}\right)

Where:

L = Spiral length from TS to SC or ST to CS

\theta = Central angle of spiral arc, L_{s} (Spiral angle); in radians

For example,

$$
\begin{aligned}
Y_{s} &=30.000\left(\frac{0.49993}{3}-\frac{0.49993^{3}}{42}+\frac{0.49993^{5}}{1320}-\frac{0.49993^{7}}{75600}+\frac{0.49993^{9}}{6894720}\right) \\
Y_{s} &=4.911\,\text{m}
\end{aligned}
$$

Deflection angle at TS from initial tangent to SC, \theta _{c} #

\theta _{c}=tan^{-1}\frac{Y_{s}}{X_{s}}

Where:

Y_{s} = Tangent offset from initial tangent of any point on Spiral

X_{s} = Tangent distance along initial tangent of any point on Spiral with reference to TS or ST

For example,

$$
\begin{aligned}
\theta_{c} &= \tan^{-1} \left( \frac{Y_{s}}{X_{s}} \right) = \tan^{-1} \left( \frac{4.911}{29.259} \right) = 9.5280 = 9^{\circ} 31′ 41”
\end{aligned}
$$

Long Chord (TS to SC or CS to ST), LC #

LC=\sqrt{\left(X_{s}^{2}+Y_{s}^{2}\right)}

Where:

Y_{s} = Tangent offset from initial tangent of any point on Spiral

X_{s} = Tangent distance along initial tangent of any point on Spiral with reference to TS or ST

For example

$$
\begin{aligned}
LC &=\sqrt{\left(29.259^{2}+4.911^{2}\right)}=29.668\,\text{m}
\end{aligned}
$$

Long Tangent of Spiral, LT #

LT=X_{s}-Y_{s}Cot\theta _{s}

Where:

Y_{s} = Tangent offset from initial tangent of any point on Spiral

X_{s} = Tangent distance along initial tangent of any point on Spiral with reference to TS or ST

\theta_{s} = Central angle of Spiral arc Ls (Spiral angle)

For example,

$$
\begin{aligned}
LT &=29.259-4.911Cot\left(28.64787749\right)=20.269\,\text {m}
\end{aligned}
$$

Short Tangent of Spiral , ST #

ST=\frac{Y_{s}}{Sin\theta _{s}}

Where:

Y_{s} = Tangent offset from initial tangent of any point on Spiral

\theta_{s} = Central angle of Spiral arc Ls (Spiral angle)

For example,

$$
\begin{aligned}
ST &=\frac{4.911}{Sin\left(28.64787749\right)}=10.244\,\text {m}
\end{aligned}
$$

Offset from initial tangent to PC of shifted circle, P #

P=Y_{s}-R_{c}\left(1-Cos\theta _{s}\right)

Where:

Y_{s} = Tangent offset from initial tangent of any point on Spiral (Y_{s} for SC or CS)

R_{c} = Radius of circular curve

\theta_{s} = Central angle of spiral arc, L_{s} (Spiral angle)

For example,

$$
\begin{aligned}
P &=4.911-30\left(1-Cos\left(28.64787749\right)\right)=1.238\,\text{m}
\end{aligned}
$$

Tangent distance from TS to PC of shifted circle, k #

k=X_{s}-R_{c}\left(Sin\theta _{s}\right)

Where:

X_{s} = Tangent distance along initial tangent of any point Spiral with reference to TS or ST

R_{c} = Radius of circular curve

\theta_{s} = Central angle of spiral arc, L_{s} (Spiral angle)

For example,

$$
\begin{aligned}
k &=29.259-30\left(Sin\left(28.64787749\right)\right)=14.876\,\text{m}\end{aligned}
$$

Total Tangent Distance, T_{s} #

T_{s}=\left(R_{c}+P\right)Tan\left(\frac{\Delta}{2}\right)+k

Where,

R_{s} = Radius of circular curve

P = Offset from initial tangent to PC of shifted circle

\Delta = Total deflection angle of curve

k = Tangent distance from TS to PC of shifted circle

For example,

$$
\begin{aligned}
T_{s} &=\left(30.000+1.238\right)Tan\left(\frac{111.297699}{2}\right)+14.876=60.581\,\text {m}
\end{aligned}
$$

External distance, E_{s} #

E_{s}=\left(R_{c}+P\right)Sec\left(\frac{\Delta}{2}\right)-R_{c}

Where:

R_{c} = Radius of circular curve

P = Offset from initial tangent to PC of shifted circle

\Delta = Total deflection angle of curve

For example:

$$
\begin{aligned}
E_{s} &=\left(30.000+1.238\right)Sec\left(\frac{111.297699}{2}\right)-30.000=25.361\,\text {m}\end{aligned}
$$

Components of Vertical Alignment #

Vertical alignment in road design refers to the longitudinal profile of the road along the designed path, displaying the changes in elevation along the alignment. Proper design of the vertical alignment, in accordance with standards, is required to provide adequate sight distance, ensuring safe stopping and overtaking conditions for vehicles while navigating the road.

Details of the curve:

VIP No.: VIP2

VIP CH.: 600

VIP Elevation: 35.91

Grade In: 1.235%

Grade Out: -2.187%

Design Speed: 30 kph

Delta Grade, \DeltaG (%) #

G=OutgoingGrade-IncomingGrade

For example:

$$
\begin{aligned}
G &= \left (-2.187 \right )-1.235 \\
G &=-3.422
\end{aligned}
$$

Crest & Sag Curves #

The vertical curves established during grade changes represent the road’s transition as it goes up or down along the terrain. There are two types of curves, crest and sag.

K & Vertical Curve Length (VCL) #

Vertical Curve Length (VCL) represents the horizontal distance required to achieve a specific change in grade, while K is the curvature coefficient required to obtain the VCL. The two components are linearly interrelated, in which an increase in K will increase the VCL as well.

This relationship can be expressed mathematically by the formula:

VCL=K\times A

Note: A represents the change in grade.

In road design, we usually aim for a longer VCL; it is equivalent to a smoother, flatter curve, which requires a higher K value. The rule of thumb is that the longer the curve, the better it is, whether it is a crest or a sag, as it provides a safe sight distance and better driving comfort.

Vertical Curve Length (m) #

VCL=k\times\Delta G

For example:

$$
\begin{aligned}
VCL &=5.000\times\left|\left(-2.187\right)-1.235 \right | \\
VCL &=5.000\times 3.422 \\
VCL &=17.11\,\text{m}
\end{aligned}
$$

K value, K #

K=\frac{VCL}{\Delta G}

For example:

$$
\begin{aligned}
K &=\frac{17.11}{\left|\left(-2.187\right)-1.235\right|} \\
K &=5.000
\end{aligned}
$$

Required Length (m) #

Length=MinK\times\Delta G

For example:

$$
\begin{aligned}
Length &=5.000\times\left|\left(-2.187\right)-1.235\right| \\
Length &=5.000\times 3.422 \\
Length &=17.11\,\text{m}
\end{aligned}
$$

Middle Ordinate, M_o #

M_{o}=\frac{\left(\Delta G\times VCL\right)}{800}

For example:

$$
\begin{aligned}
M_{o} &=\frac{\left|\left(-2.187\right)-1.235\right|\times 17.11}{800} \\
M_{o} &=\frac{58.55042}{800} \\
M_{o} &=0.073
\end{aligned}
$$