Why still no dive acceleration difference?

[BlackBerry]

Originally Posted by MadBlaster

no, that is not what I'm saying. it may provide thrust. but it is not "excess" thrust. that is the key here. excess thrust, excess thrust, excess thrust. it is not excess thrust because Crumpp posted a diagram on csp propellar that shows you can not have peak efficiency beyond Vmax. The only way to get beyond vmax and create excess thrust is to dive at the necessary angle. go back and look how he defined excess thrust. it's the difference between the two force vectors. in level flight, the force vector from gravity has no forward direction. at vmax and level flight, there is no more opportunity to create excess thrust from the prop. you have to dive to create excess thrust and acceleration.

MadBlaster, When your la7 diving at 20 degree with constant speed of 610km/h, the air drag force is completely counteracted by g*sin(20), and you turn on engine, so that you add extra thrust from engine, although the engine thrust is smaller than airdrag force, this is the excess thrust.

Both gravity---g*sin(20) and engine thrust are allied, the sum of these two, are counteracting against air drag force, since the sum of them is greater than air drag( at 610km/h), your la7 speed increases, when you reach 700km/h, air drag is quite more than g*sin(20), so a portion of engine thrust(eg 40%) will be used in completely 100% counteracting air drag. The excess thrust is from the left portion of engine thrust(60%). The more efficiency, the more excess thrust you get, understand?


If you dive at 60 degree angle reaching 700km/h, of course, your engine thrust could not 100% conteract the air drag by itself, so you need gravity--g*sin(60) to help you, if propeller efficiency is high, you need less gravity to help you, and more gravity will be used as excess thrust. if you turn off engine(efficiency=0), you need the most gravity to help, thus minimum excess thrust.

Therefore, excess thrust comes from the sum of gravity and engine thrust. Gravity and engine thrust help each other, this is teamwork, if one of them performs better, the other will have more ability to counteract air drag, Understand? The more propeller efficiency, the more sum, the faster dive.

[MadBlaster]

Originally Posted by BlackBerry

MadBlaster, When your la7 diving at 20 degree with constant speed of 610km/h, the air drag force is completely counteracted by g*sin(20), and you turn on engine, so that you add extra thrust from engine, although the engine thrust is smaller than airdrag force, this is the excess thrust.

Both gravity---g*sin(20) and engine thrust are allied, the sum of these two, are counteracting against air drag force, since the sum of them is greater than air drag( at 610km/h), your la7 speed increases, when you reach 700km/h, air drag is quite more than g*sin(20), so a portion of engine thrust(eg 40%) will be used in completely 100% counteracting air drag. The excess thrust is from the left portion of engine thrust(60%). The more efficiency, the more excess thrust you get, understand?


If you dive at 60 degree angle reaching 700km/h, of course, your engine thrust could not 100% conteract the air drag by itself, so you need gravity--g*sin(60) to help you, if propeller efficiency is high, you need less gravity to help you, and more gravity will be used as excess thrust. if you turn off engine(efficiency=0), you need the most gravity to help, thus minimum excess thrust.

Therefore, excess thrust comes from the sum of gravity and engine thrust. Gravity and engine thrust help each other, this is teamwork, if one of them performs better, the other will have more ability to counteract air drag, Understand? The more propeller efficiency, the more sum, the faster dive.

yes blackberry. this example is rigged though. if i spawn from 3000 meters and V max with the engine off, I can also create acceleration or de-cceleration simply by changing the angle of the dive and the result is a new terminal velocity/equilibrium. it's the change in the sum of the two vector forces, not simply "the sum of" the two vector forces. the thrust vector is still zero in this case.

But we have been assuming up to now the engine is on, level flight and v max.

[Crumpp]

it's the change in the sum of the two vector forces, not simply "the sum of" the two vector forces.

well said.

You understand Blackberry that is the excess force that moves the aircraft it is new equilibrium point.

The summing of the forces determines the vector of motion but the rate of change in motion along that vector is a function of the excess force.

For example, when the engine is off our thrust = 0 but our drag force remains. The summing of the forces results in a negative vector and our aircraft slows down as it seeks a new equilibrium point.

If the pilot pushes the stick forward to control the angle that he can shift a component of weight in order to counter act that drag force to maintain velocity or even speed up.

[BlackBerry]

To determine aceleration, we need the amount of excess force along our vector of motion. It is that excess force that causes the aceleration.

Right.

However, what is equilibrium point? In my opinion, the equilirium point is the point where all forces are well balanced.

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.

If you dive in a very shallow angle, for example, dive at 10 degree, the equilirium point is sth.700-800km/h which piston planes could withstand. Yes, you can reach this equilirium point.


As I said again and again, dive steeply. If you dive at 45-60% degree, you are always accelarating, you are always increasing your speed, therefore, before you pull to level flight, there is no equilirium point at all in your dive.

Since there is no equilirium point, the Newton 2nd rule tells us:

Force=acceleration*mass

Thus:

dive acceleration=(Excess Thrust in steep dive)/aircraft weight=(engine thrust + gravity along dive- air drag)/(aircraft weight.)

Imagine that there are two la7, "A" is equipped with high efficiency propeller, the other "B" with low efficiency prop, anything else being equal.

At any piont on the 45 degree dive, A always has more engine thrust than B, thus A always has bigger excess thrust than B, finally, A always has better dive acceleration than B.

[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