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- FM/DM threads (http://forum.fulqrumpublishing.com/forumdisplay.php?f=196)

- - Latest Patch - Spitfire Neg G changes (http://forum.fulqrumpublishing.com/showthread.php?t=25910)

I have no problem with accurate negG cutout, However I do have to question if a quick spike in negG would cause an instant fuel cutout. Surely the float chamber has some fuel in it when a quick spike of negG occurs? Perhaps the cutout shouldnt be so immediate, I dont know but I just think there would be a very short delay before the negG "event" occured.
I have read many pilot reports of neg G cutout when "bunting" the aircraft nose down, but I havnt read of anyone complaining of negG cutout when climbing and just trimming forward a bit to reduce the rate of climb, as modelled in CloD.
whilst on this subject surely after the fuel starve cutout there should be a overrich "flood" that causes spluttering and a cloud of black smoke (unburnt fuel) for perhaps a couple of seconds. does this happen in game ? I never fly external view but I cant say I have noticed the black smoke.

How long do you think a few ounces of fuel last's in a WWII fighter engine consuming 40-160 gallons per hour?

Even in a Lycoming O-360 A1A consuming 8-12 gph, the consumption is high enough and the cut out is for intensive purposes, instantaneous.

Certainly in a climb a float type carb will react if subjected to negative accelerations. In normal climbs, a gust or turbulence acceleration is rapid onset and very short duration so the skipping is not noticed by most pilots. If you have a digital rpm and manifold gauge or an EIS you will see it.

Bunting was a common tactic used by Bf-109 pilots to escape Spitfires on their tail in fact.

A bunt is just a maneuver involving a negative g pushover, especially in an air-to-ground attack or leveling out after a zoom climb.

Quote:

Originally Posted by whoarmongar (Post 337204)

The black smoke occurs with both the Spits and the Hurries.

Absolutely true ! The sum of external Force acting on a solid is null if the solid is at rest ;)

However when you say that you hve just pulled 5G it does not mean that you hve pulled 5g +1G of the normal gravity. It's only 5g (otherwise the G that you feel won't be the same if it was an horizontal turn or a vertical pull up). Then at level flight G =1 . Bank at 60 G=2 (see pic bellow)

By the way that where the Il2 spit was eating E ;)

PS: the neg G cut-out is just perfect : Don't make any modification for god Sake !

Quote:

Originally Posted by cheesehawk (Post 337338)

You are under 1G acceleration sitting at your computer just as I am. You are also not accelerating just as you are saying......

How can that be???? :confused:

An airplane in level flight is under 1G. The force of lift provides the centripetal force required to keep the airplane at a constant altitude.

When our butts are planted in a computer chair, it is the chair and the ground pushing back against that 1G acceleration of gravity that provides the center seeking force to keep us on the surface of this spinning ball we call Earth.

Quote:

Acceleration is change in velocity over time, which when you are sitting is 0m/s^2. By definition, when you are at rest, or traveling at a constant state, there is no acceleration.

Not when you are sitting on a round ball that is spinning....

The earth is spinning so by definition we are always accelerating...

Hence the acceleration of gravity!!! :cool:

Quote:

Any motion in a curved path represents accelerated motion, and requires a force directed toward the center of curvature of the path.

http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html

The confusion here is frame of reference. From your bodies Center of Gravity, there is a balanced force so you have no acceleration. From the Earths center of gravity, you are under a constant state of acceleration.

Back to airplanes....

G-meters are calibrated from 1G level flight....

http://www.aircraftspruce.co/catalog...htdatafc50.php

Why??

An airplane in a constant altitude turn must ALWAYS offset 1G of force in order to maintain that altitude. Most of you are savvy enough on the physics to understand that lift equals weight in 1G level flight.

When our airplane turns, it must provide the amount of force required to meet the centripetal or center seeking force required to keep us on a curved path. Fortunately in a constant altitude steady state turn, this amount of force has a fixed relationship to angle of bank.

Otherwise, the simple calculus and math required to determine reasonable prediction of airplane performance would not be so simple!! :o

At 60 degrees angle of bank, we need produce 2 times the weight of our aircraft in total force to maintain a steady state constant altitude turn. One times the weight is needed to offset the acceleration of gravity and One times the weight to provide the centripetal force required to keep us on our curved path.

From the frame of reference of the our airplanes CG and thankfully, us as the pilot, we perceive these two different acceleration vectors as one. Even though the airplane is banked our butts still want to stay down planted in the seat. Our flashlight hung on the canopy latch still hangs down pointed at the floor.

If we look at our G meter, it reads a steady 2G, just as it should and all is correct in the universe!

How does a G-meter work?

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An accelerometer is a damped mass on the end of a spring in its simplest form.

Quote:

For example, a stationary, single-axis accelerometer is adjusted so that its measuring axis is horizontal. Therefore, its output will show G Force measurement to be 0 g, and will continue to be 0 g if it is placed in a vehicle moving at a constant velocity on a level road. But if the automobile driver brakes sharply, the accelerometer will give a reading of about −0.9 g, which corresponds to a deceleration. However, the jerk due to a change in motion in the vehicle and gravity pull of the ground on the accelerometer should not be looked at as the same thing.

But if the accelerometer is turned around by 90°, so that its axis points upwards, it will calculate G Force to be +1 g upwards even though the vehicle is still stationary. Here, the accelerometer is exposed to two forces: the gravitational force and the ground reaction force of the surface it is placed on. Remember that the accelerometer can measure only the latter force, due to mechanical interaction between the accelerometer and the ground.

http://www.gforces.net/a-comprehensi...asuring-g.html

Quote:

Originally Posted by cheesehawk (Post 337338)

I most definitely am not accelerating while sitting at the computer. I may have two forces acting on me (gravity and the normal force of the chair pressing back against gravity), but am currently in a stable state of 0 acceleration.

Acceleration is change in velocity over time, which when you are sitting is 0m/s^2. By definition, when you are at rest, or traveling at a constant state, there is no acceleration.

Hi. Honestly, I do not understand what you ask. The accelerometer measures the forces, wich acting on the airplane, not the current movement of the aircraft. If the name confuses you, call it "G-meter". Its default value (standing or flying horizontally): 1. If you push the stick is reduced, when you pull increases. During inverted flight -1 ...

Quote:

Originally Posted by cheesehawk (Post 337534)

Gravity is not caused by the rotation of the earth, all mass creates gravity. The only thing close enough to us to feel is the mass of the earth, but large enough objects (i.e. the moon) also exert gravity on us. No, it has nothing to do with the moon spinning either.

Have you ever worked with accelerometer, it most definitely does not read 1g in a stationary car when turned on its side. Done here, not teaching physics if I'm not getting paid...

Ehh. Forget the cars. :)
The accelerometers measuring along the longitudinal axis in the cars, and along of the vertical axis in the aircrafts. If you working with car acc.meters, you have to do, just rotate it 90 degrees and you will see that there is 1G. In aircrafts was built that way because, the vertical forces can cross the structural load limits during tight turns or high speed maneuvers.

Quote:

Gravity is not caused by the rotation of the earth,

Nobody ever said it was caused by the rotation of the earth.

Read again what I wrote in my previous post.

Quote:

all mass creates gravity

Correct. And the distance from the masses center determines the centripetal or center seeking force required to keep you in the same spinning frame of reference.

:cool:

That is why you and your airplane lose weight when you gain altitude!! The farther from the center, the less center seeking force required. :-)

Quote:

GRAVITY was explained by Isaac Newton with a formula that says two masses M1 and M2 are attracted to each other by a Gravitational Force. The Gravitational Force, F1, is calculated by the equation below:

F1 = G x M1/R x M2 / R

Where R is the distance between the center of M1 and the center of M2 and G is the Constant of Gravitation.

If you use metric units, M! and M2 are kilograms, R is meters, G is

6.673 / 100000000000 Newton-Meters per kilogram per kilogram, and

F1 is Newtons.

CENTRIPETAL FORCE, F2, is the force that pulls on a string if you tie a rock on the end of a string and twirl the string with the rock on the end around your body. The Centripetal Force is the force that pulls on the string. A similar thing happens when a planet orbits around the sun or when our moon orbits around the Earth or when the space shuttle orbits around the Earth. The smaller object revolve around the bigger objects. The string is like gravity.

F2 = M! x V x V / R

Where V is the velocity (speed) of the rock or satellite

If you use metric units, V is meters per second

ORBITS: A less massive object M1 can orbit around a more massive object M2 if the Centripetal Force of the orbiting object is the same as the Gravitational Force between the two object. If you look at F1 and F2 above, then there is an orbit if F1 = F2. Another form of this equation is

G x M1/R x M2 / R = M1 x V x V / R

Or the speed of the satellite, M1, must be V = SQRT (G x M2 / R)

http://library.thinkquest.org/3806/ORBMECH.HTM

Cheesehawk,

VO101_Tom is trying to help you. Listen to him as he is correct.

Thank you very much for the link.
very interesting video and it was a pleasure to listen to the old man telling stories from this time.