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Influence of airspeed and engine rpm + g force equation

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Influence of airspeed and engine rpm + g force equation

Old 21st Jun 2012, 22:19
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Influence of airspeed and engine rpm + g force equation

Hi guys,

Some of you will be aware I am putting together a flight simulator, just as a learning excersize really. I am getting pretty far with the flight model now. The aerofoils are all behaving as expected.

Very early vid if anyone is interested.. ( It's come on quite a long way since then )


What I would like to do now is get the engine behaving correctly.. I have a basic force generated by the prop based on the rpm. (I will soon change that so that the prop is an actual aerofoil and generating lift much like the wings. ) What I don't understand yet though is the relationship between airspeed and RPM...well I get what happens, if you are in a steep dive / climb etc but what I am looking for is the maths that describes this relationship. Is there an equation to describe this? Or an approximation ? Or just some futher detail of the physics behind it so I can create a rough model.

Secondly g-force. How do I calculate the g's acting on my pilot? Possibly a bit simpler to answer

Thanks to anyone that has read all this and can help!

Chris
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Old 22nd Jun 2012, 08:49
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ft
 
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a = v^2/r

For the prop, I think all will become clear as you do your airfoil model. Integrate the torque from drag and lift along the blade for various conditions. Consider what the airspeed does to the AoA of the blade. If you want it simple, the prop can probably be replaced with fixed-incidence blade segment at a constant radii and still render a believable simulation.
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Old 22nd Jun 2012, 11:51
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If you've coded your flight model in anything like the usual way, you already have calculated the three components of the acceleration of the centre of mass of the aircraft. To a first approximation these can be used for the acceleration felt by the pilot. If the cockpit is substantially displaced from the centre of mass you also have to add in the corresponding moments of the angular accelerations, which your flight model should also provide.
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Old 22nd Jun 2012, 21:57
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a = v^2/r
its a bit more complicated than that!

a is acceleration due to gravity
velocity is measured in 'x' units per second, x being feet or meters
radius is the simple radius of the turn, for this calculation, in feet or meters

from the equation above, divide 'a' by 32 ft/sec/sec to get the g force.

this is an overly simplistic value, but generally is good for your purposes.

If you wanted to, you could get really technical, and figure the bank angle, degree of curvature (ie in a 2nm radius turn with a delta of 30 degrees) coupled with the weight of the pilot times the COS of the bank angle..in the full banked turn. This makes sense when you look at making a turn, with only for a few degrees of change vs a 90 degree or more.

The equation above can also be used in the vertical direction, but since that is a parabola, you would have to simplify the apex into the simple radius for the max. You will see negative G forces, which means you are reducing the weight associated with the force due to gravity.

Last edited by FlightPathOBN; 22nd Jun 2012 at 21:58.
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Old 23rd Jun 2012, 17:56
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ft
 
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FlightPathOBN,
in fact it is just as simple as that. The equation describes the acceleration (a) of an object with velocity v following a curved path with radii r in all and any conditions.

If said object is in an accelerated reference frame, such as in a gravitational field, obviously you have to account for that as well - but the equation above is all there is to it as far as acceleration due to manoeuvring goes.

I'm certain none of this is news to the OP though.

Stick with SI units and you take a lot of additional complexity out of it. 9.81 meters per second square will be just fine, even if it does vary a bit with latitude (and location). For FS work, anything between 9.5 and 10 would do I'd say - no need to confuscate matters.

Bank angle is just a means of achieving the acceleration required, but that's merely one application of the theory.

Cheers,
Fred
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Old 23rd Jun 2012, 18:38
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I think what I meant was that equation gives you acceleration or a rate of change in velocity.

If you want G force, you have to divide 'a' by 32 ft/sec/sec.
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Old 23rd Jun 2012, 19:54
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ft
 
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What is G force? It is the load factor, or nz. What is the load factor? It is acceleration. Referencing it to the normal acceleration on the surface of the earth is just a way of making the number convenient.

It is effectively a unit used to convert the g force into a quota we can easily relate to, rather than a hard-to-grasp number. "4G" tells us more than "40 m/s^2", even if it is the same data.

The question was "how to calculate the g force", not "how to transfer the g force into a convenient yardstick".

Compare it with how

speed = distance/time

is how you calculate the speed of a train.

Dividing the acceleration by the standard acceleration is the equivalent of dividing the speed by the freeway speed limit, in order to give an easier number. "It goes three times as fast as a car on the freeway".

G-force - definition of G-force by the Free Online Dictionary, Thesaurus and Encyclopedia.

"G-force
A force acting on a body as a result of acceleration or gravity, informally described in units of acceleration equal to one g. For example, a 12 pound object undergoing a g-force of 2g experiences 24 pounds of force."

My printed aviation dictionaries are in boxes at the moment.

Cheers,
Fred
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Old 23rd Jun 2012, 21:02
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Think about it....a is acceleration, correct? a change in velocity.

If you simply used v^2/r, what happens when inverted?

In an airplane, the pilot’s seat can be thought of as the hand holding the rock, the pilot as the rock. When flying straight and level at 1 g, the pilot is acted upon by the force of gravity. His weight (a downward force) is 725 newtons (163 lbf). In accordance with Newton’s third law, the plane and the seat underneath the pilot provides an equal and opposite force pushing upwards with a force of 725 N (163 lbf). This mechanical force provides the 1.0 g-force upward proper acceleration on the pilot, even though this velocity in the upward direction does not change (this is similar to the situation of a person standing on the ground, where the ground provides this force and this g-force).
If the pilot were suddenly to pull back on the stick and make his plane accelerate upwards at 9.8 m/s2, the total g‑force on his body is 2 g, half of which comes from the seat pushing the pilot to resist gravity, and half from the seat pushing the pilot to cause his upward acceleration—a change in velocity which also is a proper acceleration because it also differs from a free fall trajectory. Considered in the frame of reference of the plane his body is now generating a force of 1,450 N (330 lbf) downwards into his seat and the seat is simultaneously pushing upwards with an equal force of 1,450 N (330 lbf).

For the pilot, horizontal axis and vertical axis G forces must also be considered as they have a much different effect and tolerances in the body.

Last edited by FlightPathOBN; 23rd Jun 2012 at 21:08.
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Old 23rd Jun 2012, 21:20
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ft
 
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What you are describing in a lot of words is what is usually described, in rather fewer words, as an accelerated reference frame.

Originally Posted by ft, above
If said object is in an accelerated reference frame, such as in a gravitational field, obviously you have to account for that as well - but the equation above is all there is to it as far as acceleration due to manoeuvring goes.
If you break it down further, it's actually all

F = m * a,

with a = v^2/r being the above applied over a curved trajectory of fixed radii.

It is simple. If it becomes complex, it is usually through thinking too much.

Cheers,
/Fred
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