Why still no dive acceleration difference?

[Crumpp]

You dive in 45 degree with 110% WEP, your speed is always increasing, BEFORE you reach the so called " equilibrium point", you lost your wings.

Try il2 4.11m, use Tempest, P51D, P47, fw190d,bf109K, La7, to dive in 45 degree with full throttle, you can NEVER find a equilirium point where the speed stops increase.Could you? Could you? Why? The equilirium point is sth 1300km/h! Those planes will explode before reaching 1300km/h, believe it or not. They are not supersonic a/c.

What are you talking about here?

Do you need a method to estimate the velocity of the new equilibrium point given the increase in forces?

[Crumpp]

there is no equilirium point at all in your dive.

Sure there is blackberry, you just have not reached it yet.

Your equilibrium point velocity is found by converting the Vmax TAS to EAS. We are going to change altitude and we don't have to constantly mess with density effects.

Now we can use the relationship our parasitic drag component to find our new velocity. You already know the Cdo of the design.

So using the relationship of parasitic drag force and velocity:

Dp2 = Dp1(V2/V1)^2

Re-arrange to solve for velocity:

At 100 KEAS our airplane produces 5000 lbs of parasitic drag. At what velocity will it produce 10000lbs of parasitic drag?

100 KEAS * SQRT(10000 / 5000) = 141 KEAS

If we suddenly gained 5000lbs of thrust (5000lbs plus 5000 lbs) our new equilbrium velocity would be 141 KEAS.

[BlackBerry]

Originally Posted by Crumpp

Sure there is blackberry, you just have not reached it yet.

Your equilibrium point velocity is found by converting the Vmax TAS to EAS. We are going to change altitude and we don't have to constantly mess with density effects.

Now we can use the relationship our parasitic drag component to find our new velocity. You already know the Cdo of the design.

So using the relationship of parasitic drag force and velocity:

Dp2 = Dp1(V2/V1)^2

Re-arrange to solve for velocity:

At 100 KEAS our airplane produces 5000 lbs of parasitic drag. At what velocity will it produce 10000lbs of parasitic drag?

100 KEAS * SQRT(10000 / 5000) = 141 KEAS

If we suddenly gained 5000lbs of thrust (5000lbs plus 5000 lbs) our new equilbrium velocity would be 141 KEAS.

When P51 at Vmax=700km/h, what's the air drag force? 1500lbs?

When P51 dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1200km/h!

When Bf109K dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1100km/h!

When P47D dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1300km/h!

When TempestMKV dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1200km/h!


When Dora dive in 45 angle with full throttle, what's the next equilibrium piont speed?

1100km/h!

Since those planes never reach so called the next equilibrium piont, the equilibrium is totally useless in analysis of 45 degree dive acceleration.Since you have to pull your aircraft out of 45 degree angle far before reaching the equilirium point, you are always accelerating in dive, you always have excess thrust during dive, higher efficiency propeller always provides higher thrust.

The sum of gravity and engine thrust is always bigger than air drag force.


I post the picture again, do you understand me?

q.jpg

[MadBlaster]

blackberry, the sim would be very boring if all the forces were balanced in equilibrium. there would be nothing to do! I can only think of one time that occurs, when your on the ground and the engine is off and the wind is zero. equilibrium point is not something that has to manifest to exist. obviously, there are constraints that may make it impossible to achieve (e.g., max dive speed). It is the process of moving from set of conditions to another. Dynamic verses Static. This is the concept that is being pointed out.

[BlackBerry]

Originally Posted by MadBlaster

blackberry, the sim would be very boring if all the forces were balanced in equilibrium. there would be nothing to do! I can only think of one time that occurs, when your on the ground and the engine is off and the wind is zero. equilibrium point is not something that has to manifest to exist. obviously, there are constraints that may make it impossible to achieve (e.g., max dive speed). It is the process of moving from set of conditions to another. Dynamic verses Static. This is the concept that is being pointed out.

Yes，in 45 degree dive, aircrafts are always unbalanced , aircrafts are always increasing speed, always accelerating.

This is a dynamics process, in this process, if my la7 throttle is 100% while your la7 throttle is 60%, I can always outdive you when moving to the equilibrium point which impossible for real flight.

Why? My la7 100% engine produce more thrust than your 60% engine in 45 degree dive, so I have more excess thrust during the whole dive, finally, I can get better dive acceleration than you. understand?

Even if you turn off fuselage damage in difficulty setting, both of us could achieved "equilibrium" speed:

I, with throttle 100%, equilibrium speed is 1000km/h
You, with throttle 60%, equilibrium speed is 950km/h

I can still pull away from you because my engine give me more thrust. Look this picture, when both of us reach equilibrium point, my air drag force is higher than you, this is to say, my speed is bigger than yours.
qq.jpg

Now , do you still believe that engine could not produce excess thrust during high speed>Vmax dive?

[MadBlaster]

Sorry, I need more information because the example keeps changing. What is the speed of the two planes before they go into the dive? Is the one at 100% throttle already at Vmax? Or are they both at Vmax because the the one at 60% dove down from higher altitude previously? Honestly, this example doesn't seem good one. The information is sketchy. Are you trying to trick me?

[BlackBerry]

Originally Posted by MadBlaster

Sorry, I need more information because the example keeps changing. What is the speed of the two planes before they go into the dive? Is the one at 100% throttle already at Vmax? Or are they both at Vmax because the the one at 60% dove down from higher altitude previously? Honestly, this example doesn't seem good one. The information is sketchy. Are you trying to trick me?

Imagine there are two tempestMKV named 'X' and 'Y' respectively.

X is equipped with a good CSP, while Y is equipped with a bad CSP. Everything else being equal. both a/c weight equal. drag coefficient equal, engine is equal ......

Within the envelop, within Vmax, both good propeller and bad propeller share same efficiency, they performs identically. good prop=bad prop=85% efficiency

But when dive into 0.6-0.7Mach which is out of a/c envelop, good propeller=80% efficiency, bad propeller=60% efficiency.

At first, both X and Y use 100% throttle in level flight SIDE BY SIDE, because they share same efficiency within evelop, both speeds are Vmax.

Then, both begin to dive at 45 angle, and after a while, both speed are entering 0.6-0.7 Mach, suddenly, "Y" CSP lose efficiency to 60%, while "X" efficiency is still 80%. That is to say, "X" engine could provide more thrust.

Question: Will "X" begin to pull away from "Y" from now?