Again, constant acceleration formulars are wrong in calculating airplane's dive.

Do you mean when the airplane reach 624fps(424mph), the acceleration is zero?


OMG, why don't you open il2 4.11m and choose any piston aircraft, spawn at 7500m altitude and dive at 45 degree with full throttle?

I promise your aircraft will lose wings sooner or later. You could NEVER reach the equilibium point where the speed stops increase.

Crumpp, plz run il2 4.11m and try some tests, it's very easy for you to find the truth.

This is the expression for the average acceleration:

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

It is all in the formula, blackberry.

Think about it. Do you know what the acceleration would be if just gravity is acting alone on the airplane?

32.2ft/s^2.....

Do you think our acceleration is going to greater or smaller than the acceleration of gravity at the beginning of the dive when we have the most excess thrust?

GREATER THAN

Why is our average acceleration lower than that of gravity???

20.98 ft/s^2 < 32.2ft/s^2......

Because there is an aerodynamic braking force acting on the aircraft that lowers the average acceleration.

Yes the formulation requires constant acceleration but that does not mean it does not accurately describe the motion of the aircraft.

For this airplane under these conditions...YES.

It is called terminal velocity.

g=32.2ft/s^2=9.8m/s^2

At the beginning of 45 degree dive, dive acceleration is greater than g due to bigger engine thrust and smaller air braking force, and at the end of dive, acceleration is smaller than g due to smaller engine thrust and bigger air braking force.

But we need to calculate acceleration very accurate. In a 15 seconds long 45 degree dive, if my average acceleration is slightly than yours, for example, is 3ft/s^2 or 0.9m/s^2 more than yours. I'll be 50km/h faster than you in the end. That's a hugh advantage.

Do you remember the 60 degree dive between P47D and fe190G in 1943 summer? At the end of dive, P47D had a much greater speed than fw190G, sth. like 50km/h difference is huge enough.

I have a question for you:
At 1st second, you begin to 45 degree dive,
At 2nd second, your speed increase a little, acceleration is greater than g.
at 3rd second, your speed increase a little more, and acceleration is still greater than g.

.....

at 20th second, you reach the equilibrium point, and acceleration is ZERO. Your speed remains same.
at 21st second, your speed is same as 20th second.
at 22nd second, hour speed is same as 20th second.
......

How about the 19th second? what's your acceleration? It should be smaller than g, but I need more accurate data.

Is the acceleration at 19th second nearly zero or around 0.5*g?

Look at what the aircraft are placarded for in terms of dive speeds.

The aerodynamic forces will tell you the FW-190A can outdive the P47. Just look at the sea level speeds and power required. At sea level, EAS = TAS and EAS is the speed the airplane always feels.

The P47 generates ~272 THP more than the FW-190A8 to travel ~20mph slower. It takes a lot of power to push that big heavy P47 through the air.

However, the relationship of thrust and drag is not the primary limiting factor in a dive for these airplanes. Dynamic pressure limits and mach limits tend to set the speed limits in WWII fighters.

In the case of the FW-190 vs P 47, the FW-190 is limited to ~466mph at low altitude while the P47 is limited to 500 mph at low altitude.

Those placard limits are not set arbitrarily nor is there any wiggle room or safety factor. A Focke Wulf pilot exceeding 466 mph is taking a huge risk he will not survive the dive. There are plenty of incidents of FW-190 pilots diving straight into the dirt barrier because mach effects made the elevator control ineffective. There are also incidents of the pilot turning the aircraft to confetti by aggressive use of the trim to recover.

In a dive to Vne, the P47 will always gain ~34mph over the Focke Wulf FW-190A.

Come on, Achilles and the Tortoise story is very simple.

Sum of time=1+0.5+0.25+0.125+........ is limited, will never bigger than 2.
of course, Achilles can't over take tortoise WITHIN 2 seconds because tortoise is some distance ahead of Achilles in the beginning.

In ancient time, people didn't know the property of "infinite series", so they were puzzled. They drew the conclusion that Achilles would never overtake tortoise because there are infinite numbers of time series in the sum, We modern people will laugh at them because the sum of infinite numbers is limited to "2".

My story, is totally different, from 1st to 20th seconds, a 45 degree diving a/c acceleration changes from g(usually higher) to 0. During the dive process, a/c moves to equilibium point. The acceleration is always decreasing. If you treat it as a constant acceleration motion, you'll make a mistake.

I am a little disappointed that we havn't known each other's opinion after many posts.


Crumpp, P47 Vne=500mph, Fw190's Vne=466mph, the difference is 34mph, so if both of them dive steeply and pull to level flight at Vne speed, P47 is always 34mph faster than fw190. That' very simple, all of us know it. However, that is "terminal speed difference", not dive acceleration difference.

Although X plane's Vne is more than Y plane's, it is possible that X is outdived by Y WITHIN Y's Vne speed. The test in Italy 1943 between P47D an Fw190G is very useful to demonstrate this.

That is my explanation:

g=9.8 m/s^2=32.2 ft/s^2

When both Fw190G and P47 dive in 65 degree angle, their acceleration along path(65 degree) is calculated in this formular:

Acceleration=g*sin(65)+g*(thrust/weight)-air braking force/weight

Fw190A has better thrust/weight, P47 has less braking force/weight, but at low speed, fw190's advantage is more profound, so P47 was outdived by fw190g INITIALLY, although P47's Vne (500mph)higher than fw190(466mph). This perfectly demonstrates that terminal speed and dive acceleration are totally two different concepts.

In later dive period, P47 passed fw190, both arrived 3000ft but P47 has much greater speed.Both speeds of them are well BELOW 466mph. Furthermore, P47 got higher speed BEFORE reaching 3000ft, isn't it? Don't forget fw190 was ahead of P47 after initial dive.

For example, when fw190 reached 8000ft altitude, P47 was at 8500 ft, so P47 was outdived by fw190 initially.On what altitude did P47 get same speed as fw190? Perpahps at 5000ft, at that time, fw190 was still ahead of P47, so fw190 probably at 4500ft. BEFORE P47 got same speed as fw190, P47's acceleration was better than fw190, otherwise, how could P47 get same speed? Don't forget P47 was slower than fw190 after initial dive. So P47 time line is below:

Altitude
@8500ft, was slower and after fw190
@6500ft, began to get higher acceleration than fw190, but still slower than fw190
@5000ft, get same speed as fw190, but still after fw190
@4000ft, faster than fw190, but still after fw190
@3000ft, passed fw190 with higher speed and higher acceleration. At this time, both P47 and Fw190 speed are BELOW 466mph.

@1000ft(if don't pull to level), ahead of fw190, faster than fw190, higher acceleration than fw190 and the distance between them are enlargening.

@-5000ft(deep valley! Air density =SL), P47 still has a little acceleration, while fw190 reaches equilibium point at 1200km/h (if firm enough). Distance between them is big.

@-7000ft (deeper valley, air density=SL), p47 reaches equilirium point with 1400km/h(if firm enough)and fw190 is still 1200km/h, distance between them is bigger.

@-10000ft(deepest valley,air density=SL), p47 is still 1400km/h, fw190 is still 1200km/h, distance between them is even bigger and increasing for ever.


if @1000ft level flight(pull his stick to avoid loss), ahead of fw190, faster than fw190, and LOWER SLOW DOWN RATE than fw190, distance between them are enlargening.


The most interesting is the level flight out of envelop, above Vmax. When both P47 and Fw190 fly at Vmax at sea level, P47's 7ton weight couldn't himself at all, and P47 has a bigger wing area while same 2000HP engine with fw190, so P47's Vmax is less than fw190.

But if both of them level fly above Vmax, the 7 tons weight helps P47 a lot because the heavier, the more ability to retain high speed above Vmax.

Acceleration=g*(thrust/weight)-air braking force/weight

Acceleration is always negative when you are slow down from 800km/h to 600km/h on the deck, and it usually takes a whole minute/60 seconds to slow down to 600km/h(Vmax).

Thus if P47's big 4-blade propeller efficiency is 20% higher than fw190's small 3-blade at the speed range from 680 to 800km/h IAS(out of envelop), fw190 will suffer a lot of ENERGY BLEEDING.

We all know bf109k4 has a excellent climb rate over La7, it usually takes k4 a whole minute to establish 500 m altitude advantage if both have same energy in the beginning. P47/P51/Tempest could also establish 500 m higher altitude if they use high speed boom and zoom tactics-----their own method! AS EFFECTIVE AS BF109K4.

It is obvious for all il2 players if 109K4' s climb rate is degraded to la7 level. People will shout:"Sh*t, where is 109K4's climb advantage? how can I do a energy fighting?" Unfortunately, if il2 neglects the propeller efficiency difference/sharp drops between various aircrafts at high speed(680-800km/h), P51P47Tempest will lose half of their energy fight tactics effect------energy-saving high speed boom and zoom, and not many players will notice it, many players don't know that P51P47Tempest are tigers WITHOUT teeth in il2. The teeth had been taken by someone who, for some reason, didn't pay enough attention to high Mach number/compressibility effect which leads to efficiency decrease of a CSP.

Blackberry, it's complex calculus algorithm. But in computer, this is handled/simulated by optimizing the algorithm every few milliseconds. He already said, he doesn't have access to the flight model. What exactly to you expect him to do? His formula describe at a moment in time when forces are max. He would have to expand the formula from Tzero to Tn when terminal velocity is reached.

Really, I shouldn't talk about math or physics. I had to copy off my buddy's papers to get through high school. Just a heads up.;)

Ah, what the hell.

T=time. If you take your formula below and integrate it over T(intial) until T(terminal), where T(terminal) is time when terminal velocity is reached and T(initial) is when forces are at maximum, I think that gives you your answer. But I make no promises.

Acceleration=g*sin(65)+g*(thrust/weight)-air braking force/weight