After taking off, a helicopter climbs in a straight line at a constant angle β. Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d, β, θ and θ.

Question

After taking off, a helicopter climbs in a straight line at a constant angle [latex]\beta[/latex]. Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d , [latex]\beta, heta[/latex] and [latex]\dot { heta }[/latex].

Accepted Answer

From the diagram

[latex]\frac{r}{\sin \left(180^{\circ}-\beta ight)}=\frac{d}{\sin (\beta- heta)}[/latex]

or [latex]\quad d \sin \beta=r(\sin \beta \cos heta-\cos \beta \sin heta)[/latex]

or [latex]\quad r=d \frac{ an \beta}{ an \beta \cos heta-\sin heta}[/latex]

Then

[latex]\begin{aligned}\dot{r} & =d an \beta \frac{-(- an \beta \sin heta-\cos heta)}{( an \beta \cos heta-\sin heta)^{2}} \dot{ heta} \& =d \dot{ heta} an \beta \frac{ an \beta \sin heta+\cos heta}{( an \beta \cos heta-\sin heta)^{2}}\end{aligned}[/latex]

From the diagram

[latex] u_{r}= u \cos (\beta- heta) \quad ext { where } \quad u_{r}=\dot{r}[/latex]

Then

[latex]\begin{aligned}d \dot{ heta} an \beta \frac{ an \beta \sin heta+\cos heta}{( an \beta \cos heta-\sin heta)^{2}} & = u(\cos \beta \cos heta+\sin \beta \sin heta) \& = u \cos \beta( an \beta \sin heta+\cos heta)\end{aligned}[/latex]

or [latex]\quad\quad\quad\quad u=\frac{d \dot{ heta} an \beta \sec \beta}{( an \beta \cos heta-\sin heta)^{2}}\blacktriangleleft[/latex]

Alternative solution.

We have [latex]\quad u^{2}= u_{r}^{2}+ u_{ heta}^{2}=(\dot{r})^{2}+(r \dot{ heta})^{2}[/latex]

Using the expressions for r and [latex]\dot{r}[/latex] from above

[latex]\begin{aligned} u & =\left[d \dot{ heta} an \beta \frac{ an \beta \sin heta+\cos heta}{( an \beta \cos heta-\sin heta)^{2}} ight]^{2} \ ext{or}\quad\quad u & =\pm \frac{d \dot{ heta} an \beta}{( an \beta \cos heta-\sin heta)}\left[\frac{( an \beta \sin heta+\cos heta)^{2}}{( an \beta \cos heta-\sin heta)^{2}}+1 ight]^{1 / 2} \& =\pm \frac{d \dot{ heta} an \beta}{( an \beta \cos heta-\sin heta)}\left[\frac{ an ^{2} \beta+1}{( an \beta \cos heta-\sin heta)^{2}} ight]^{1 / 2}\end{aligned}[/latex]

Note that as [latex] heta[/latex] increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen.

[latex] u=\frac{d \dot{ heta} an \beta \sec \beta}{( an \beta \cos heta-\sin heta)^{2}}\blacktriangleleft[/latex]

Question 11.168: After taking off, a helicopter climbs in a straight line at ...

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