# Flighty energy Posted by

A problem: An aeroplane that weighs 1000kg is flying at 3500ft with a speed of 100kts. What speed would the aeroplane gain on descending to 3000ft, assuming NO power change, no configuration/flaps changes and the air friction loss is 15%?

Bonus questions:

1. Does it really depend on the mass of the aeroplane given?
2. Would the flight path affect the answer – e.g. steep dive or steady descend?

This problem illustrates the energy state of flight and how the altitude and speed of the aeroplane are related. While we know it from ab initio flight training and intuitively feel it every time we fly, I just wanted to show here how this works through maths for the nerds here.

To understand that, we need to recall the school physics law of conservation of energy. It states that energy cannot be created or destroyed, although it can be changed from one form to another.

For an aeroplane in flight, we can consider 2 main types of energy:

• Potential energy Ep, due to the gravitation of Earth, which depends on the mass of the aeroplane (m), the altitude (h) of the flight and the gravitational constant (g): Ep=m*g*h
• Kinetic energy Ek, due to the momentum of the movement, which depends again on the mass of the aeroplane (m) and squared velocity (v) of it: Ek=1/2*m*v^2

(The units of the above formulas are meters for altitude, kilograms for mass and meters per second for velocity).

What the last conservation of the energy means is that the sum of these energies is a constant (with a given power setting and ignoring the aerodynamic drag), i.e.: Ep+Ek=C, where C is some constant value.

What it else means is that a change in one energy causes an equal in value but opposite in sign change of the other energy. E.g. descent (negative change in potential energy) will cause a change in velocity (increase in kinetic energy).

So, here is some basic math:

• Change of potential energy due to change from altitude h1 to altitude h2: dEp=m*g*(h1-h2)
• Change of kinetic energy due to change from velocity v1 to velocity v2: dEk=1/2*m*(v1^2 – v2^2)
• The law of conservation of energy means dEp=-dEk, as stated above.

Now, just to make it slightly more realistic, if we consider an effect of the aerodynamic drag (D) as consuming energy of the flying aeroplane and use the formulas above and the appropriate unit conversions, we get:

• The change of energy due to descent dEp=10009.81(1066.8-914.4)=1495044 (joules)
• Assuming 15% would be lost to aerodynamic drag, the effective change in kinetic energy dK=-dP(1-0.15)=1270787.4 (joules)
• Solving for v2 (new speed) from the known values, we get: v2=SquareRootOf(dK2/m+v1^2)=1270787.4*2/1000+(51.4)^2=72.03m/s or ~140kts

It works!