Why still no dive acceleration difference?

[Crumpp]

Since those planes never reach so called the next equilibrium piont, the equilibrium is totally useless in analysis of 45 degree dive acceleration.

Airplanes certainly do reach their equilibrium point in a dive.

However the most common restriction to dive performance is dynamic pressure limits <flutter> and mach limits.

Completely irrelevant though as the equilibrium point estabilishes the rate of aceleration in the dive.

You can debate it all day long but it does not change the fact it is how performance is predicted.

Maybe it is already accounted for in the modeling of the drag force?

The excess force is what drives the aircraft to equilibrium. It is all accounted for in the formulation.

so this interplay between opposing thrust and drag vectors gets masked over by weight/dive vector. am i on the right track?

The interplay is covered in the derivative.

You start out with the maximum force which is the moment in transitioning that a component of weight has shifted to thrust and the propeller thrust is still at level flight velocity. That is the maximum excess force you will have available.

Let's look at the rectilinear motion equations and solve both a zoom climb problem and a dive problem using the same airplane at the same entry speeds.

We will end our zoom at Vy or best rate of climb speed and end our dive at the equilibrium point.

We are going to just simplify the drag and thrust values to illustrate the mechanics of solving the problem.

Characteristics of our Airplane:

Weight 9000lbs
Thrust in lbs = 1000lbs
Drag in lbs = 500

Zoom climb from 300mph to Vy at a 45 degree angle:
The airplane will move to equilibrium

Entry speed = 300mph = 441fps
Zoom Angle 45 degrees
Vy = 150mph = 220.5fps

Zoom height:

Sum the forces on the flight path -

9000lbs * sin 45 = 6364lbs
1000lbs – 500lbs - 6364lbs = 5864lbs

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 5864lb/279.5lb-s^2/ft
a = 20.98 ft/s^2

s = (V1^2 – V2^2 ) / 2a

s = (441^2 – 220.5^2)/(2 * 20.98ft/s^2) = 3476.18 ft

3476.18 ft * sin 45 = 2458 ft

Now let's dive under the same conditions:

Characteristics of our Airplane:

Weight 9000lbs
Thrust in lbs = 1000lbs
Drag in lbs = 500

Zoom climb from 300mph to Vy at a 45 degree angle:

Entry speed 300mph = 441fps
Angle of Dive 45 degrees

We need a ballpark of our equilibrium speed. A quick method is to use the relationship of Parasitic drag. It is the major drag component at high speed. More detailed analysis will give better results but this is accurate within 10%. 10% is acceptable for climb/dive performance.

441(SQRT 1000/500) = 624fps

At 624 fps, the acceleration about the CG will be zero.

Dive:

Sum the forces on the flight path -

9000lbs * sin 45 = 6364lbs
1000lbs – 500lbs + 6364lbs = 6864lbs

a = F/m

m = 9000lbs/32.2 = 279.5 lb-s^2/ft
a= 6864lb/279.5lb-s^2/ft
a = 24.6 ft/s^2

s = (V1^2 – V2^2 ) / 2a

s = (441^2 – 624^2)/(2 * 24.6ft/s^2) = -3961ft (negative is a vector direction)

3961ft * sin 45 = 2801 ft of altitude lost!!