This won't work.
Lets look at this rationally. Anyone who has ridden a plane knows that it takes off and then goes to a certain altitude, and during that time accelerates to speed
vc. This takes time
t. During time
t, v is increasing from
v0 to
vc at some rate, which we don't know. But lets say for the sake of argument- that it's always 30 min to get to altitude and its always 30 min to land, and during that hour speed is always less than
vc. Lets say this distance traveled during this approximately and hour is constant (which its not, but for ease of calculation) and call it
X. Otherwise we'll have to deal with integrals, and I don't want to do that here.
So in a 10 hour flight, 9 hours are at cruising speed
vc, but in a 4 hour flight only 3 hours are at
vc. This is 90% and 75% of the flight respectively. Which means when you compare 2 non-similar lengths, the ratio will be severely skewed.
Next, lets take short flights vs. long flights. Flights of different lengths usually have different planes, which have different cruising speeds.
So if you want to make a map using only this information and not the speed themselves, then you have to have an adjustment factor,
I, to adjust the time by the ratio of speeds between Plane 1 and Plane 2 (
v1/v2). It's easiest to pick a single base plane and set that as
I=1 and compare all the rest to that. So now your times will be comprable.
So, a barely reasonable formula might be
((t-1)*vc)*I+X.
Now you also have to account for prevailing wind speed, which will change the speed of the plane-
+w for tailwinds and
-w for headwinds. You have to adjust the speed
w to account for the fact that it might not be perfectly parallel to the plane, which we can find by taking the angle between the direction of travel and the direction of wind-
Theta.
This gives
((t-1)*(vc)+w cos (theta))*I+X Then, you also have to account for general human elements, that might change a flight- usually the planes will go a little faster or slower to make sure that they arrive in a particular time window so that there is a gate available. Your presumption that planes
always travel the perfect speed is wrong. My car has a listed cruising speed of 60mph, but I'll push it to 80mph if I'm late, and slow down to 40mph if the weather is bad. Planes also have different cruising speeds at different altitudes. This is because the oxygen content in the air changes and the engine has to maintain sufficient flow through it, etc. etc. So all of these things together add an error of lets say- extrememly conservatively 10%.
So to find the distance between two seperate line segments since we don't believe in maps or modern technology such as GPS-you want the ratio between two distances which would be sufficient to draw an unscaled map- you can calculate:
((t1-1*h)*(v1c)+w1 cos (theta1))*I1+X/ ((t2-1h)*(v2c)+w2 cos (theta2))*I2+X.
It might seem that
X would cancel, but it won't. We also need to adjust that by
I because
I is the term that adjusts based on
vc. We can adjust
X by dividing by
I (since we will cover distance X faster if
vc is faster. But we also need to adjust X by an additional factor-
h that takes into account 2 things- the altitude we are at before we hit
vc, and the time it takes to get there-
((t1-1h)*(v1c)+w1 cos (theta1))*I1+(X*(h2/h1/I) /
((t2-1h)*(v2c)+w2 cos (theta2))*I2+(X*(h1/h2/I) +/- 10%. This would give you a minimimally accurate ratio that you can calculate using the information you have at hand. You can
not do any less, because you are talking about making a
map afterall. Not just guesstimating how far it is. You could do this all in Excel. Only from flight aware you also need to grab the type of plane, and maybe the windspeed from there or somewhere else. The +/- 10% is also critical.
You also have to remove the additional distance that the planes fly because they
think they are travelling along a curved earth. I've got to get back to work, but you can find the math here-
https://math.stackexchange.com/questions/830413/calculating-the-arc-length-of-a-circle-segmentBasically if you fly from Paris to NYC you have to figure out what that distance would be on a round earth- and divide the arc length by the line segment length to come up with a ratio. This would be a correction factor. This correction factor, A, grows in significance the longer the distance- to a maximum of
pi/2*d where
d is the diameter of a supposedly round earth. Since you are using flights that go across most the world, or at least across the country, this becomes pretty important. You can find it by multiplying the change in Latitude squared + the change in latitude squared to the 1/2 power (pythagoras' theorom).
(ΔLat2 + ΔLong2)(1/2)- thats your round Earth distance in Lat/Long- convert that to miles (since our speeds are in miles)- and then divide that by
d. Thats your Round Earth distance traveled. To Calculate the Flat Earth Equivalent is tougher. I'd have to think about that. but it would be:
FEE/RDE * ( ((t1-1h)*(v1c)+w1 cos (theta1))*I1+(X*(h2/h1/I) /
((t2-1h)*(v2c)+w2 cos (theta2))*I2+(X*(h1/h2/I) ) +/- 10%. So, most of the eq. is:
Flat Earth Eq. / ((ΔLat2 + ΔLong2)(1/2)) * ( ((t1-1h)*(v1c)+w1 cos (theta1))*I1+(X*(h2/h1/I) /
((t2-1h)*(v2c)+w2 cos (theta2))*I2+(X*(h1/h2/I) ) +/- 10%. Just leaving you one piece to find yourself. You should be able to do it from there.