FLIGHT PATH ANGLE
Here is my take on the slope problem.
I do these quite often but only use the Trig function (right angled triangle) and the SIN as this is the angle for slant range DME distance.
Here are the two formulas I use
1. For gradient, Gradient ÷ 6076 = SIN of angle
6076 x SIN = Gradient
Grad ÷ SIN = 6076 (# of feet in a NM)
2. For Height and Distance
Height ÷ Distance NM = Gradient
Distance x Gradient = Height AGL
Gradient x height = Distance NM
In the right angled triangle, the opposite side (side opposite the angle) represents height above the ground
The hypotenuse represents the slant range DME distance and the adjacent represents the flat (map) ground distance. When you fly over the DME station (in a jet) the distance goes down to about 5 nm then starts going up again. That’s because of the slant range distance, since you are never less than 5 nm (31000 ft) from the station.
Let’s check out at the approach to Birmingham Airport. Notice the 3.28 slope.
crossing baskn at 2300 ft 4.7 nm from the threshold, TDZ = 644 ft (2300 - 644 = 1656 feet to loose). If we divide the heigh by the distance we get a grad of 1656 down for each nm forward. Now had I been doing that approach, here is what I would have done instead.
I would have used the 3.3 deg slope 5.8%, and 350 ft per nm. That would have had me crossing Colig at 5120 ft ASL (4480 AGL) and using the Flight Path Angle Computer which all Airbus planes have, I would have easily and safely executed the approach and be alive today. Pilots must understand that there is NO requirement to cross or be at the published minimum altitude. Those altitudes are “NOT BELOW” altitudes, nothing else.
So, Colig, 12.8 nm, 5120 asl, and each NM after, 350 ft lower,all the way to 2.8 nm at 1624 ft and 1.8 nm at 1274 ft. Pilots need to use more math in the cockpit to stay alive.
Hope some of you will take this valuable info and put it into practice.