For jets, how can we show that 1/2*ρ*V^2 is equivalent to 1481.3*δ*M^2?

In Roskam's book, we find the following relationship given for climb performance in jet aircraft:

Thrust requested = $D$ = $C_d \cdot (1/2 \cdot \rho \cdot V^2) \cdot S = C_d \cdot (1481.3 \cdot \delta \cdot M^2) \cdot S$

How can we show that $(1/2 \cdot \rho \cdot V^2)$ is equivalent to $(1481.3 \cdot \delta \cdot M^2)$?

SOURCE: Author: Jan Roskam
Title: Airplane Aerodynamics and Performance
Paragraph "9.2 CLIMB AND DRIFT-DOWN PERFORMANCE OF JET AIRPLANES"

• See 14.3.4 and how equation 14.27 is obtained. – mins Jul 25 '16 at 11:13

Using the Mach number definition $M=\frac{V}{\sqrt{\gamma RT}}$, the ideal gas equation $p=\rho RT$ and the atmospheric pressure ratio $\delta=\frac{p}{p_0}$:
$$\frac{1}{2} \rho V^2 = \frac{1}{2} \rho \cdot \gamma RT \frac{V^2}{\gamma RT} = \frac{1}{2} \gamma p \cdot M^2 = \frac{\gamma \cdot p_0}{2} \cdot \delta \cdot M^2$$
With $p_0=2116.2 \text{lbs}/\text{ft}^2$ and $\gamma = 1.4$ you get the desired result.