How Does Side-Slip Affect Lift Calculation in Flight Simulations?

[janger]

I'm writing a flight sim for use in Cinema 4D, an animation package that has a built-in physics engine. It's not supposed to be totally realistic, and I've got most things working fairly well. But one thing I don't understand is how to calculate lift correctly when the relative wind isn't coming directly from the "forward" direction.

What I mean is, say the plane points along the z-axis, and the wing's leading edge points along the x-axis, but the wind also has some x-component due to the aircraft side-slipping. Most books state something like: "lift is perpendicular to the direction of the relative wind". But I don't see how the x-component can have an impact on the lift calculation since it doesn't really "flow over" the wing. It's coming from the side. Are you supposed to zero out or ignore the x-component, and use the resulting vector as the relative wind? So a velocity, v = <100, -10, 90> would become v = <0, -10, 90>, and use this in the airspeed/lift equations?

 

[rcgldr]

The cross wind component should not create a significant amount of lift. Assuming this was a conventional aircraft with some amount of dihedral (the wings tilted up a bit for stability), a relative cross wind would tend to roll the plane away from the crosswind (self correction of the side slip).

 

[janger]

Ok but the thing is if I zero out the relative side wind component and use the resulting vector in the lift calculations, I can make the aircraft do barrel rolls etc fine. But if I use the entire relative wind vector, weird things happen. Things like the lift force seemingly going negative when the wings are almost pointing straight up and down. I don't know if this is due to "the gimbal effect" or what.

I'm obviously confused on how to calculate the lift direction. The magnitude comes from the lift equation and that's fine. To calculate the direction I've been finding the cross product between the local x-axis (along the wing leading edge) and the relative wind vector. Is this correct? If so, then why can I only make it work when I zero out the side wind component?

Would it be possible for you to outline how to do this correctly, sort of step-by-step? I thought there was an old post here with some pseudo-code but I can't find it.

EDIT: Here's some code from the Grit Engine:

pvel=body:getLocalVelocity(pcent, false)
speedx = dot(pvel, body.worldOrientation * vector3(0,-1,0))
speedy = dot(pvel, body.worldOrientation * instance.actionVectors[s])
pvel = speedx * (body.worldOrientation * vector3(0,-1,0)) + speedy * (body.worldOrientation * instance.actionVectors[s])
speed2 = speedx*speedx + speedy*speedy
aoa = -dot(norm(pvel), body.worldOrientation * instance.actionVectors[s])

lift = speed2 * instance.liftCurves[s][aoa]

Isn't this just using the wind components that are perpendicular to the wing's leading edge and ignoring the side component when calculating the lift magnitude? If so, then which vector do I multiply it with to get the final lift vector?

 

[rcgldr]

The lift related to angle of attack and speed. A simple approximation would use the linear range of angle of attack so lift would be the product of angle of attack time speed² times a coefficient. If you knew the "polar"s for the air foil, you could implement a more accurate equation. Since the elevator controls the angle of attack, for a given speed, lift should remain constant, even in a barrel roll, as long as the elevator control input (which controls the pitch attitude and angle of attack) remains the same.

Also, the lift force is "up" (or "down") relative to the air craft's orientation, not the relative wind. The exception to this is knife edge type flight where the fuselage is yawed a lot (lots of rudder input) and acts as a wing. In radio control aerobatic models and perhaps some full scale planes, it's possible to do loops all in knife edge flight.

 

[janger]

rcgldr, thanks for your help. But I've spent the last few days reading through books and checking out some code and am no closer to understanding this. And it seems another member you've helped in the past is having the same troubles understanding this concept.

The thing is, I get how lift works and how to calculate it when the relative wind is coming from the front, ie perpendicular to the wing's leading edge. That's a no brainer. What I don't get is what to do when the wind has some side component. I'm talking about during aerobatic maneuvers etc, like your knife edge example. In these cases I'm unsure how to calculate the lift magnitude and direction correctly.

A thought experiment:
Using the same coordinate system I mentioned in the OP, where the fuselage points along the z-axis, the wings leading edge points along the x-axis, and y is up.

Let's say the plane is flying with no yaw, and the relative velocity is

v1 = <0, -10, 100>

. Using the dot product this gives an AoA of 5.7 deg, or 0.099 rad. So plugged into the lift coefficient equation:

CL1 = 2.pi.AoA = 2.pi * 0.099 = 0.622

Now, the pilot enters some aerobatic maneuver which puts the plane in a position where the relative velocity is now

v2 = <30, -10, 95.4>

This is the same speed as v1 above. The angle between the wing chord (z-axis) and the relative velocity, which is how the books define AoA, is now:

AoA = acos( DotProduct( <0,0,1>, vnorm(<30, -10, 95.4>) ) ) = 0.32 radians

And CL is now:

CL2 = 2.pi * 0.32 = 2.01

Wooah! When plugged into the lift equation, the second calculation is giving approx 3 times the lift, from the same airspeed! And it's flying "diagonally" into the wind. Do you see why I'm confused now?

So, where is my mistake here? Or, are you supposed to project all vectors onto a plane which lies parallel to the wing chord, and then do all the calculations using those projected vectors? Which was really my original question.

And, is it absolutely true that no matter what, the direction of lift is perpendicular to the actual relative velocity vector, ie perp. to v1 and v2 above? Even if you do zero out the side wind component for the intermediate calculations?

 

[rcgldr]

To keep this simple, assume the lift from the wings is only related to the wind that flows opposite of the direction of the aircraft, and that "lift" from the fuselage is only related to a cross wind. The third type of flow, an up draft or down draft, also perpendicular to the direction of the air craft, could be treated as drag.

 

[janger]

Haha, that's what I'm trying to do rcgldr. Keep it simple :)

"the wind that flows opposite of the direction of the aircraft" is obviously what I showed in the calculations above. It's the opposite of the relative velocity. But this doesn't help me any. As I showed, using the relative velocity gives a larger lift magnitude when the relative wind is at an angle to the direction the plane (wing) is pointing. How can this be? What is wrong with my calculations?

And the most important question: In which direction is the lift vector? Is it perpendicular to the relative wind, no matter what? Or is it perpendicular to the relative wind ignoring any side component?

The reason I'm having trouble understanding this is practically every introductory book shows a 2d image of a cross-section of a wing, with the relative wind in the same geometric plane. But planes fly in a 3d world, don't they :).


You see, I can visualize several ways of doing this, and I don't know which is correct (or why this is so hard for me to understand):

1. Just use all the vectors as is. But as shown in the calculations above, this leads to a higher CL value when the wind has some side component (AoA is larger). This method makes my simulation unstable.

2. Project the relative velocity vector onto a plane that the wing chord axis lies in. Essentially this converts the calculations into the 2D equivalents shown in the introductory texts, ie all the vectors, including the resulting lift vector, are in the same plane. So the lift vector will never have any sideways component (relative to the wing). However this goes against the "lift is perpendicular to the relative velocity". But it makes my sim more stable.

2a. Even if you do step 2, still use the actual velocity vector (all x, y, & z components) to calculate speed and the lift vector. This also makes the sim unstable.

So basically, I just need someone to show me the correct calculations using the values I defined in my previous post, in pseudo-code, c# code, pascal, I don't care. Just so I can see how this side wind component should be handled. And how the resulting lift force and it's direction is also computed.

Thanks,
Dave

 

[rcgldr]

The angle of attack is independent of the aircrafts direction. If you assume wings are oriented so they generate no lift if the plane is pointed in the direction of flight, then the AOA is the angle between the aircraft's direction and the aircraft's orientation, along a plane that is perpendicular to the wings. For example if the plane is pointed in the direction of the z axis, {0, 0, speed}, then the angle of attack would be measured along the y-z plane. If the plane was generating lift, then it would be oriented "upwards". The velocity vector would be {0, 0, speed} and the orientation vector would be {0, up, forward}. The angle of attack would be tan^-1(up / forward).

 

[janger]

Right, so you're saying you basically zero out any side wind component when calculating AoA?

From my previous calculations:
The wing's chord points along the z-axis = <0,0,1>,
v(rel) = <30, -10, 95.4>

Therefore, AoA is the angle between <0,0,1> and <0, -10, 95.4>. And not between <0,0,1> and <30, -10, 95.4>
Is this correct, or do I still not understand?

If that's correct, then in which direction is the lift vector? Perpendicular to <0, -10, 95.4>?
Or perpendicular to <30, -10, 95.4>? In other words, just like the AoA is measured in the y-z plane, is the lift vector calculated in that same plane, or is it still perpendicular to the relative velocity (no matter which direction the wing is pointing)?

If you can answer these questions, I think I can stop bugging you :) And if this is correct about calculating everything in the same y-z plane, then I've wasted my time as well as yours trying to fix something that wasn't broken, ie I was doing it correctly in the first place, lol :)

 

[rcgldr]

I'm not sure you understand my point. The path of the aircraft is independent of it's orientation. It could be possible for the aircraft to be pointed upwards but falling downwards in what's called a tail slide in aerobatics.

You're going to need to maintain a separate vector (a second set of x, y, z parameters) for the aircraft's orientation in addition to the aircraft's velocity vector. To calculate lift and drag, you'll need to convert the aircraft's velocity vector into one that is relative to the aircraft's orientation, in order to calculate lift and drag. This is assuming a zero wind condition, or that your'e using the air as a frame of reference. If there's a wind condition, you can just add that in at the end of your calculations to determine the aircraft's path relative to the ground.

 

[janger]

I'm not sure you understand my point.

No that's ok. We obviously don't understand one another. :)

My problem isn't with the basics here. I have both global and local matrices and all that stuff. The "math" part is not giving me a headache. That's all fine. It's simply which angles/vectors/vector components etc to use in the calculations. Basically I'm calculating everything in "aircraft coordinates". That is, relative wind is relative wind, relative to the aircraft's orientation. Ok? I convert it back to world coordinates before actually applying the forces. This works easier in C4D.

To calculate lift and drag, you'll need to convert the aircraft's velocity vector into one that is relative to the aircraft's orientation, in order to calculate lift and drag.

This! This is the bit I'm having trouble getting right!

So here's another scenario: this time we'll do it in a wind tunnel so the plane is fixed. Hopefully this will avoid confusion. So, once again same coordinate system - fuselage pointing in the z-axis, wing LE along x-axis, y is up. The plane is mounted pointing upstream, with zero pitch on the wing. Yep, this airfoil creates lift at 0 AoA. You turn the wind tunnel on with an airspeed of 50 m/s. This gives a relative velocity (in aircraft coordinates!) of v1 = <0,0,50>. So the lift vector points in a direction perpendicular to the relative wind. And also perpendicular to the wing's leading edge (x-axis). This would be along the aircraft's up, or y-axis, correct?

Now, grab that plane and turn it 45 degrees anti-clockwise when looking from the top. But keep the pitch, roll exactly the same. Turn on the wind tunnel again with 50m/s. Hey, it's still the same airspeed, isn't it? And still the same AoA. But the rel velocity vector is now v2 = <35.4, 0, 35.4>.

Ok, so how has rotating the aircraft affected the lift force? How has it effected the lift direction? Is the lift vector now perpendicular to <35.4, 0, 35.4> ? Because the books state "lift is perpendicular to the relative wind". And that vector is the relative wind (well, opposite direction, but you know what I mean).

 

[rcgldr]

This airfoil creates lift at 0 AoA. You turn the wind tunnel on with an airspeed of 50 m/s. This gives a relative velocity (in aircraft coordinates!) of v1 = <0,0,50>. So the lift vector points in a direction perpendicular to the relative wind.

The normal definition for lift is the force perpendicular to the direction of the aircraft with respect to the air, using the air as a frame of reference. Using the aircraft as a frame of reference, the relative wind could be used as an alternative.

Now, grab that plane and turn it 45 degrees anti-clockwise when looking from the top. So how has rotating the aircraft affected the lift force?

Using a simplified view of this, the upwards force from the wings has been reduced, so that the upwards force is now 70.7% ( sin(45 degrees) ) of what is was before. There is now a side force to the left from the fuselage, also perpendicular to the relative wind. You'll need to define a coefficient of lift versus yaw AOA for the fuselage. At a 45 degree angle, the fuselage is probably in a stalled state (increased drag, reduced side force). Note that both the upwards force (y axis) and the side force (x axis) are considered "lift" since they are both perpendicular to the relative wind, in spite of the fact that they are also perpendicular to each other. There's also a drag force in the direction of the relative wind (z axis).

 

[sophiecentaur]

I have only come into this half way through but wouldn't it be likely that the lift would be modified by Cos(angle plane is rotated wrt its forward velocity) as the length of wing hitting the air would be shorter by that amount? It would be easy to try that.

Also, how does dihedral come into this? I reckon that could make a huge difference to lift if the plane isn't going in the direction it's pointing: lots more lift, all of a sudden, as it would be like banking.

 

[rcgldr]

Cos(angle plane is rotated wrt its forward velocity) as the length of wing hitting the air would be shorter by that amount?

I was assuming angle between wing and relative wind to be 90 degress when the plane is moving straight ahead, which was probably a bad idea, but it's too late to edit that post now.

If the angle is more properly defined as planes yaw angle versus direction vector (or relative wind), then it would be cosine as you mentioned.

Also, how does dihedral come into this? I reckon that could make a huge difference to lift if the plane isn't going in the direction it's pointing: lots more lift, all of a sudden, as it would be like banking.

Dihedral generates a roll torque if a crosswind is experienced. There are some model air craft that have no ailerons and rely on dihedral to bank the plane based on rudder inputs. A swept back wing also provides dihedral effect, and is used on B-52's which have wing tip spoilerons as opposed to true ailerons, but this limits their roll control authority.

 

[janger]

I have only come into this half way through but wouldn't it be likely that the lift would be modified by Cos(angle plane is rotated wrt its forward velocity) as the length of wing hitting the air would be shorter by that amount?

This is what I have been asking. They say a picture is worth a thousand words. Is the following the correct way to calculate all the entities (explanation follows):

I forgot to include the axis labels: x is red, y is green and z is blue.

The yellow plane is the y-z plane, relative to aircraft (wing) coordinates.
Vr is the "true" relative velocity. Once again, relative to aircraft coordinates.
Vp is Vr projected onto the y-z plane. As seen in the image, if Vr = <x,y,z>, then Vp = <0, y,z>. You basically zero out the x-component. Actually what I was asking in the OP

So,
1. Is the angle marked AoA the correct one?
2. Do I use the length of Vp as the wind speed in the lift equation, and not Vr?
3. Does the lift vector also lie in the y-z plane as shown? ie. is it perpendicular to Vp, and not perp. to Vr?

The x-component which was zeroed out would be added back in at the end and basically only affects ground speed.

Have I got this right? Because it's what I was doing in the first place.

 

[rcgldr]

Is the angle marked AoA the correct one?

It's correct for the AoA of the wings, but you don't show the AoA for the fuselage, which would be the angle between the projection of Vp and Vr onto a plane parallel to the x-y plane but going through the center of the aircraft. It would be better if the z axis went thorugh the center of the aircraft.

Do I use the length of Vp as the wind speed in the lift equation

The length of Vp is the wind speed for the up force equation, not the total lift (which would also include side forces from the fuselage).

Does the lift vector also lie in the y-z plane as shown?

Only the "upwards" component of lift lies in the y-z plane. There's also a "side" force from the fuselage, in a plane parallel to the x-y plane going through the center of pressure of the fuselage, and this side force is the other component of lift.

The drag force would be in the direction of -Vr.

 

[janger]

Sorry rcgldr, that diagram is just for the lift produced from the wing. I know there are other forces to calculate. But the important thing was that I had this right since the same equations are used for the tail etc. And yeah I have the drag force worked out too.

It would be better if the z axis went through the center of the aircraft.

The axes are just to show orientation of the plane, and what coordinates I'm using. Since I think most books have the aircraft pointing in the x-axis rather than the z-axis. It's a "see what I was talking about" thing, not a "this is what I'm doing". In the sim I'm actually calculating lift at several points of each surface, not just one.

But do you see what I was talking about "zeroing out the x-component", and "projecting the relative velocity onto the geometric plane"?

 

[rcgldr]

Sorry rcgldr, that diagram is just for the lift produced from the wing. I know there are other forces to calculate. But the important thing was that I had this right since the same equations are used for the tail etc. And yeah I have the drag force worked out too.

In that case the diagram looks ok for the simplified calculation of upwards force from the wing. Normally the yaw angle on an aircraft is very small except for knife edge flight.