pelmet wrote: ↑Mon Aug 22, 2022 4:10 pm
It is an interesting question. Probably best for the wind tunnel and computers.
Dont need a computer and wind tunnel for this question, it's fairly strait forward and has an answer achieved with just a little bit of calculus, and if you are not handy with calculus we can come up with an intuitive description of what happens. We look at the two aircraft as being in a high drag state (operating faster than best glide) and in an optimum drag state (operating at best glide). We also know that drag is a function of velocity squared, ie doubling the speed does not double the drag, it goes up by a factor of 4.
If the aircraft has a very small delta between starting velocity and best glide, it wont make much difference, which is the case for your typical GA aircraft. But lets consider the high performance, or even extreme performance case where the aircraft begins at a velocity 4x that of the best glide speed.
In the strait ahead scenario, the aircraft simply bleeds off kinetic energy strait ahead, the rate at which that energy is shedding is a function of drag, ie, velocity squared all the way from point A to point B. There is no increase in potential energy along the way, ie, it does not climb, it simply sheds kinetic energy along the way.
Now look at the case of the aircraft doing the zoom climb. During the zoom climb it will shed the same amount of kinetic energy as the first aircraft, ending up at the top of the zoom with a given amount of kinetic energy, but also with a whole bunch of potential energy. From there it will trade potential energy for kinetic energy at best glide speed. Because of the lower optimum glide speed, for each meter moved forward, this aircraft will use up less of it's stored energy than the one operating at a much higher speed, this is simply the V squared factor in the drag equation,and if you want to get pendantic about it, the air density factor in the drag equation is slightly lower in this scenario as well.
You can do all the fancy math (I did this math back in the days when I was in school on this subject), it involves a significant amount of calculus because velocities are not constant and air density is not constant, but in the end you will find as aircraft 2 arrives at point B, it will have significantly more energy than aircraft 1, and that energy will be in the form of potential energy, ie altitude. It will take longer to get there, but it will be significantly higher when it does.
But the mistake a lot of folks are making in this thread, they are considering the point of engine failure as the start point of the calculation, that is wrong. The start point of the calculation must begin at the point the pilot recognizes the failure, and begins to react. In your typical light GA aircraft, if there was any excess kinetic energy at the point of failure, it's likely already burned off at the point of recognition, the delta is just not that great. But with a high performance aircraft operating at 2 or 3 times best glide speed at the point of failure, there will be LOTS of excess kinetic energy available, and the best way to save it up for use later, is to convert it to altitude and decelerate the aircraft into a state that burns it off more efficiently.
For perspective, there is another scenario that boils down to the exact same math in terms of conservation of energy. You are short of gas, do you push the throttle in to 'get there faster', or do you pull the throttle back, fly at best range speed, take longer to get there but burn less gas in the process ? In an airplane, gas and altitude are the same thing, potential energy, just stored differently.