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<p class="MsoNormal"><span style="mso-fareast-language:EN-US">It looks like your inputs are in degrees (of longitude and latitude), so then you need to do your calculations accordingly (you can’t switch to radians, unless you first convert your inputs into
radians).<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">To convert degrees of latitude travelled into a distance on the ground (assuming no change in longitude, and that what you are measuring does not go up over one of the poles and back down the other
side), multiply by the circumference of the earth over 360 (or equivalently, by a quarter of the circumference of the earth over 90). You can see that this is right if you consider an object travelling along say the Greenwich meridian from the equator to
the North Pole. It will have travelled 90 degrees of latitude, and multiplying this by a quarter of the circumference of the earth over 90 will give the result: a quarter of the circumference of the earth, which indeed is the distance travelled.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">Of course, to convert from the diameter of the earth to the circumference, multiply the diameter by pi.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">To convert degrees of longitude
<b>at the equator</b> to latitude, do the same calculation. It does not work away from the equator. You will see on a globe with degrees of longitude and latitude marked on it, that the degrees of latitude never get closer together, but the degrees of longitude
converge at the poles. So if you travel at the equator from say 0 degrees to 90 degrees West, you will have travelled a quarter of the circumference of the earth, but if you are just a foot from the North Pole when you do it, you can do it in a single step.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">So away from the equator, you need to take not the circumference of the earth, but the circumference of the line of latitude that you are on (lines of latitude of course being circles), which is
of course two*pi*radius of the line of latitude. Visualisation of a right angle triangle with one of the short sides being the radius of the circle described by the line of latitude in question, and the other short side going from the centre of that circle
to the centre of the earth (and you will see that the hypotenuse of the triangle extends from the centre of the earth to the surface and therefore is equal to the radius of the earth), shows that the radius of the line of latitude is equal to the radius of
the earth times the sine of the latitude.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">So the formula for the distance travelled by an object moving along a line of latitude through a certain amount of ‘degrees’ of longitude with be equal to the number of degrees of longitude travelled
times the circumference of the earth times the sine of the latitude divided by 360.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">If you have a change in both longitude and latitude, then having calculated the latitude distance and the longitude distance (this will not be exact because you have to decide what latitude to use
in your formula, when in fact the latitude has changed, but provided the latitude has not changed by much it will be close enough), you can then use Pythagoras’ theorem, as you have done, to calculate the total distance (which will be the hypotenuse of the
right angle triangle formed whose (shorter) sides are the change in longitude and the change in latitude). Note here that Pythagoras’ theorem does not work on a sphere (to see that this is so, think of a ‘triangle’ on the earth where one side lies along the
equator from 0 latitude to 90 degrees West, one side runs from the equator to the North Pole along the Greenwich meridian, and one side runs from the equator to the North Pole along the 90 degrees West meridian: all angles of this triangle are right angles,
so any of the three sides can be considered the ‘hypotenuse’, but all sides are equal so the triangle does not accord with Pythagoras’ theorem). However, the shorter the distance, the more nearly the sides will accord with Pythagoras.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">Also note that there are various other approximations in play here, such as the earth being considered a perfect sphere (which it is not) and that the drone is travelling close enough to the surface
of the earth that the extra distance is has to travel by virtue of being higher up is not appreciable (think of a space-ship orbiting the earth very far away from the earth and you will see that is has to travel much larger distances to cover the same number
of degrees on the ground below it).<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">So in summary, you could use as a (hopefully) reasonable approximation the formula: Distance = (pi*D / 360 )*Square-root [ {V2 – V1}^2 + { sin(V1)*(H2 – H1) }^2 ]
<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">Where V2, H2 are the final latitude and longitude respectively, V1 and H1 are the initial latitude and longitude, all in degrees, and D is the diameter of the earth. Note that I have arbitrarily
used sin(V1) in the formula rather than sin(V2) or even the sine of the average of V1 and V2 (which might be best), on the assumption that the change in latitude will be so small that for your purposes it does not matter which is used. (Note that all of this
might not be very accurate in calculating the distance between the Tower of London and the Arc de Triomphe, for reasons I have already mentioned, and depending of course on how accurate you require it to be, but I presume that with the drone itself, you will
be taking readings far more frequently than the time it takes to fly that distance, assuming it is even capable of it.)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">I did ‘A’ level maths, though I don’t think the above goes beyond ‘O’ level.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">If anyone spots an error in my formula, not doubt they will speak up!<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">I’ve had to delete the rest of the thread because I got a message saying that my message was too big.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-language:EN-US">Jason<o:p></o:p></span></p>
<p class="MsoNormal"><o:p> </o:p></p>
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