# How does free space path loss change with receiver's antenna's height

I have a question which seems like an easy one, but I couldn't find anything on Internet to be 100% sure.

The free space path loss equation is:

$$FSPL=20\text{log}(d)+20\text{log}(f)+20\text{log}(\frac{4\pi}{c})$$

Now, if the receiving antenna is at the same height as the transmitting antenna, $$d$$ is just the horizontal distance between the two. But if we now move the receiving antenna up and down (without increasing $$d$$), the straight line between the two antennas is this:

and the path loss equation becomes:

$$FSPL=20\text{log}\bigg(\sqrt{(h_{T_x}-h_{T_r})^2+d^2}\bigg)+20\text{log} (f)+20\text{log}(\frac{4\pi}{c})$$

is this correct? Maths seems to agree, but I don't know if I am not missing somthing obvious here. Thanks.

Your equation is correct for your definition of $$d$$. However, your definition of $$d$$ seems incorrect. $$d$$ is supposed to the be distance between the transmitter and the receiver, that is the distance on a straight line from the transmitter to the receiver. In your diagram you have (I think incorrectly) defined $$d$$ as the horizontal distance on some plane, presumably the surface of the Earth, that the transmitter and receiver are not both on. Remember, this is free space that we're talking about, so it doesn't make sense to talk about horizontal distance on a reference plane.