# On FSX flight lesson on climbs and descents, what does "free ride" mean?

The following is from a lesson on Microsoft Flight Simulator X regarding climbs and descents:

While engine power moves a car uphill, gravity pulls it down. Without your foot on the accelerator, the car's downward speed is determined by the steepness of the hill it's descending. The steeper the hill, the faster it goes. If the hill becomes shallower, then the speed decreases. If the hill becomes too shallow, then some power is necessary to maintain sufficient forward speed.

Airplanes can also move downhill without power.

Just lower the nose, and you'll get what appears to be a free ride (it isn't, but let's not get into that).

I cannot seem to understand what the author meant by the last sentence. Is the author trying to say that descents are not "free" as in descents are not "powered" by the weight of the aircraft and some form of thrust is needed?

I am rather new to the concepts of the physics of flying and may be missing out on something. Hopefully, someone is able to point out what I am not catching. Thanks so much!

If I remember correctly most of those FSX lessons were done by Rod Machado. He’s a good CFI but has a tendency to speak tongue-in-cheek about things. What he means by that is that by lowering the nose you can maintain the same airspeed but use less power to do it i.e. getting a free ride out of it. In truth what’s happening is that the airplane is compensating for the reduction of thrust by exchanging potential energy for additional kinetic energy, and it does this by descending. It’s not exactly a “free ride” per se - energy is conserved after all - but the effect is similar to a car rolling down a hill whereby the driver does not need to give the engine gas to maintain a given speed.

Aircraft can trade altitude for speed and vice versa. Probably that's what the author meant (but only he knows for sure).

I find the quoted text misleading for people without prior knowledge of aircraft dynamics. Lowering the nose does not necessarily descend the plane (although most of the time it does).

Please get a copy of "Stick and Rudder" by Wolfgang Langewiesche. It's an excellent book, eye opener, and fun to read.

One of the main points of this book is that - opposite to beginner's intuition - the "elevator" (stick) controls the speed of the plane along its path, and the engine thrust controls the vertical speed (ascend/descend) of the plane. Actually it's a bit more complicated than this. That's why I recommend the book.

A try to clarify your comment, although I'm afraid it will only confuse you more: Descending means loosing altitude. Altitude means energy. That energy has to go somewhere. Nose down means trading that energy gained from altitude into kinetic energy (speed). Since the plane must not exceed the maximum (or required) speed, this kinetic energy must be dissipated. That happens through drag. Drag increases with speed, and speed will increase when pointing the nose down, thus dissipating energy faster. Thrust only increases the kinetic energy (speed) of the plane, instead of dissipating it.

• Thanks so much for your reply! Is it logical for me to interpret it as such: Descents require more than just the weight of the aircraft i.e. some form of thrust is needed for the descent (either from engine power or trading altitude for thrust). Hence, it is not "free"? Dec 6, 2019 at 9:46
• No, disagree. You don't need thrust to descend.
– bogl
Dec 6, 2019 at 9:55
• In a nose up descent, thrust is needed isn't it? Dec 6, 2019 at 10:35
• No. What makes you think so?
– bogl
Dec 6, 2019 at 11:57
• BTW, your last comment together with the image would be a nice question on its own!
– bogl
Dec 6, 2019 at 13:58