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ProfRob
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I know nothing about the engineering aspect, but you get a doppler shift for a radio signal whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

As has been pointed out in comments below, this is obviously a very small effect when dealing with non-relativistic velocities. Whilst the principle doppler effect is of order $v/c$, the transverse doppler effect is of order $v^2/c^2$. Hopefuly you will get further comments or answers on how you could feasiblyfeasibly do this for a RC aircraft using some alternate techniqueusing some alternate technique (I think the effect is just too small for any detection technology available), but the above is what the doppler effect can tell you (which is what you asked).

I know nothing about the engineering aspect, but you get a doppler shift for a radio signal whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

As has been pointed out in comments below, this is obviously a very small effect when dealing with non-relativistic velocities. Whilst the principle doppler effect is of order $v/c$, the transverse doppler effect is of order $v^2/c^2$. Hopefuly you will get further comments or answers on how you could feasibly do this for a RC aircraft using some alternate technique, but the above is what the doppler effect can tell you (which is what you asked).

I know nothing about the engineering aspect, but you get a doppler shift for a radio signal whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

As has been pointed out in comments below, this is obviously a very small effect when dealing with non-relativistic velocities. Whilst the principle doppler effect is of order $v/c$, the transverse doppler effect is of order $v^2/c^2$. Hopefuly you will get further comments or answers on how you could feasibly do this for a RC aircraft using some alternate technique (I think the effect is just too small for any detection technology available), but the above is what the doppler effect can tell you (which is what you asked).

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ProfRob
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I know nothing about the engineering aspect, but you get a doppler shift for a radio signal whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

As has been pointed out in comments below, this is obviously a very small effect when dealing with non-relativistic velocities. Whilst the principle doppler effect is of order $v/c$, the transverse doppler effect is of order $v^2/c^2$. Hopefuly you will get further comments or answers on how you could feasibly do this for a RC aircraft using some alternate technique, but the above is what the doppler effect can tell you (which is what you asked).

I know nothing about the engineering aspect, but you get a doppler shift for a radio signal whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

I know nothing about the engineering aspect, but you get a doppler shift for a radio signal whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

As has been pointed out in comments below, this is obviously a very small effect when dealing with non-relativistic velocities. Whilst the principle doppler effect is of order $v/c$, the transverse doppler effect is of order $v^2/c^2$. Hopefuly you will get further comments or answers on how you could feasibly do this for a RC aircraft using some alternate technique, but the above is what the doppler effect can tell you (which is what you asked).

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ProfRob
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I know nothing about the engineering aspect, but you get a doppler shift, for a radio signal whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

I know nothing about the engineering aspect, but you get a doppler shift, whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

I know nothing about the engineering aspect, but you get a doppler shift for a radio signal whatever the angle that the plane's velocity is to the line of sight.

If you are at the centre of a circular motion, then the shift can be attributable to the time dilation experienced by the plane and the observed frequency is decreased by a factor $(1-v^2/c^2)^{1/2}$.

The more general expression would be $$ f = \frac{f_0 (1-v^2/c^2)^{1/2}}{1 + (v/c)\cos\theta},$$ where $\theta$ is the angle that the velocity of the source makes (in the reference frame of the observer) with the line between source and observer ($\theta <\pi/2$ means the source is getting further away).

See https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect

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ProfRob
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ProfRob
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