[ds4719]

Hello everyone

I have a special topic that I wish to fully understand with your help. I have done some research however I have not found any conclusive results or articles that accurately describe the condition.The original thread that sparked this investigation is given here:

http://www.fsae.com/forums/showthrea...th-wings/page2
Starting on Pg 2 with the main description on Pg 4

The topic is:
The Center of Pressure (CoP) is used to help define the sum of all aerodynamic forces and moments about a singular point for a given system (in our case the entire race car). The common theory is that this point can give us an accurate description of the forces acting on the car when based from a certain origin point, usually the Center of Gravity (CoG). With this information, we can calculate the distance of the CoP to the CoG, which results in an understanding of the expected stability of the car.

According to the link and discussion above, this description of the CoP is an incomplete picture of the actual forces acting on the car. This is best said through quotes from the thread.

“The distributed air pressure forces of aerodynamics DO NOT have a "centre of pressure".

The resultant of the pressure forces is best described with a "force screw" (aka a "wrench"). This "dual vector" quantity acts along a "line of action", and NOT at any specific point, or "centre".”

To my understanding, the Aerodynamic forces are acting along a “line of action” (LoA), which is similar to a roll center, and is constantly changing based on the dynamic motion of the car.
The original CoP theory is shown in the picture below, the Instantaneous CoP would be measured using the Forces and Moments result control in Simscale with the origin point at the CoG position. The value of the CoP is the result of a moment calculated from the CoG origin point, which will tell us the distance and force acting upon the CoG due to the summation of all aerodynamic devices. While this is useful in telling us overall stability, it does not give us information on the forces at the locations that really matter, which is directly at the contact patches of the tires.



Now going back to the topic of resultant pressure forces being described as a force screw. The force screw theory is described best by quoting the thread

“Consider a rectangular coordinate system with its origin fixed to some point on our car (could be the CG, but doesn't really matter). X is forward, Y to the left, and Z upward. The car has been through the wind tunnel and we get the following "aero forces and moments", measured at the origin, for some wind speed and car orientation. (I'll use "couples", labelled with Ts, because in the long run it makes more sense.)

Forces ...................... Couples..................
================================
Fx(0) = -300N, ....... Tx(0) = +600Nm.
Fy(0) = +400N, ...... Ty(0) = +800Nm.
Fz(0) = -1,200N, .... Tz(0) = +2,400Nm.

We vectorially sum the forces to get a resultant force vector Fr(0) that points mostly downward, but also slightly backward and to the left. The magnitude of this resultant force is 1,300N (exactly!). Likewise, we sum the couples and get a resultant couple vector Tr(0) of 2,600Nm, which points mainly upward and slightly forward and to the left (using "RH screw rule").


We now draw the "arrow pairs", or vector duals Fr(P) + Tr(P), at Z=0 and Z=10. Then on an X-Y grid from -6 to 6 in steps of 3. Forces are red, couples blue, with their tails on the crosshairs, which represent the various points P. The couple vector at the origin Tr(0) is barely visible, hiding behind the Z-axis.



Fr(0) + Tr(0) is one way of specifying the "dual vector quantity" that fully describes the aero forces. So we draw some kind of 3-D picture of this with the two resultant vectors shown as arrows, with lengths proportional to some scale, and their tails at the origin.

But we also know that we can "calculate forces and moments" at any point. The origin was arbitrarily chosen, and is not special in any way. So it is time for some research. We start calculating new Frs and Trs at a whole lot of other points P.

The force Frs are simply moved to P unchanged (see note below). But new couple Trs must be calculated. The equations required are similar to Moop's above, but with opposite signs (going in opposite direction?), and we have to include the original couples.

========================
Tx(P) = Tx(0) + (-Fz*Py + Fy*Pz)
Ty(P) = Ty(0) + (+Fz*Px - Fx*Pz)
Tz(P) = Tz(0) + (+Fx*Py - Fy*Px)
=========================

Example 1. Point P(x,y,z) = (1,0,0)
Tx(P) = 6 + (12*0 + 4*0) = +6Nm.
Ty(P) = 8 + (-12*1 + 3*0) = -4Nm.
Tz(P) = 24 + (3*0 - 4*1) = +20Nm.

Example 2. Point P(x,y,z) = (10,10,10)
Tx(P) = 6 + (120 + 40) = +166Nm.
Ty(P) = 8 + (-120 - 30) = -142Nm.
Tz(P) = 24 + (-30 - 40) = -46Nm.

Technical notes. In the above, we are adding equal and opposite copies Fr(P) + -Fr(P), of the original resultant force Fr(0), at each point P. We can do this because together they sum to zero, so we are adding nothing. Fr(P) becomes the "moved" Fr(0), while the original Fr(0) + -Fr(P) becomes a new couple dTr(P) to be added to Tr(0).

Also forces are "sliding vectors" (they can slide anywhere along their LoA) and couples are "free vectors" (can go anywhere), but we put both Fr(P) and Tr(P) at P, to show that these are the "forces and moments calculated at P".

Importantly, we only need one of these "vector duals" to give a complete specification of the force system. However, we calculate lots of them to get a better view of the big picture.
~~~o0o~~~

This video accurately describes this process through solving determinants and cross products.
https://www.youtube.com/watch?v=BaM7OCEm3G0

Basically I understand this as - when we select our forces and moments about a point (as said above this can be the CoG but not necessary), the resultant force and moment given about this point “P” is not along the line of action. The origin in this Force screw Fr(0) and Tr(0) picture above is also not along the line of action. To find this line of action, more is described below:

The maelstrom of swirling couple vectors clearly show a "helicoidal field" with a calm central axis. This axis is parallel to all the force vectors, and passes a short distance in front of and to the right of the origin. The further away from this axis, the more furiously the couple vectors swirl, like a tornado with wind speeds that increase without limit away from the core.

Looking closer, we see that this particular force screw has a left-hand thread. This was actually quite obvious from the beginning, because the force and couple vectors at the origin were pointing in approximately opposite directions.

At the central axis the couple vector is at its minimum magnitude, and is collinear with the force vector. Away from the axis the couple vectors always have this same component in the direction of the axis, but they gain increasingly large circumferential components at increasing radii. The circumferential component's magnitude is simply |Fr| x radius.

All the above suggests that one way to specify any particular force screw is by giving four numbers to locate its central axis, together with two numbers for its force and couple magnitudes (with suitable definition of which is the +ve direction, etc.). So six numbers in all, with the two magnitudes being invariant to the chosen reference frame. Or we can use Plucker's coordinates, but that is another story...

Interestingly, we see that any force system, no matter how complicated (eg. composed of a squillion little air pressure forces), can always be reduced to nothing more than the application of a screwdriver. Just a push and a twist along some axis. Or a wrench = "~n. a forceful twist or pull".

But, perhaps even more interestingly, if we had to build a structure to resist this "force system", then the most sensible place to put it is in "the eye of the storm". Here the forces (well, really only the couples) are at their smallest. If we put the structure far off to the side of the axis, then there are huge "bending moments" that must be resisted. This reason alone should be enough to encourage engineers to "always find the central axis".
~~~o0o~~~

Finally, there is a direct one-step method of finding the location of the screw axis, when given the force and couple vectors at a single point. The direction of the axis is parallel to the resultant force vector Fr, and the axis' radial distance away from the point is found by considering the circumferential couple component, as briefly described four paragraphs up.


Relating this to our FSAE car, If we select the origin point for the forces and moments calculation, the Force Screw would look something like the picture below. The selection of the CoG is irrelevant as the Line of action can be calculated from any point selected. The way to find the line of action is described above as multiplying the force component (Fr)by the Radius.



To continue with my attempt at understanding how this works, The following picture shows how the Force and Couple Vectors all have a determinate that point directly towards the line of action for every point P selected in this Force Screw. I believe this is what he means by saying “Away from the axis the couple vectors always have this same component in the direction of the axis, but they gain increasingly large circumferential components at increasing radii.”

Each Couple vector Tr(P) at some point P, always have one component making up this vector pointing in the direction of the axis.



This can be much easier described by zooming in on the origin, adding hypothetical points 1&2, and the line of action shown below. All determinate vectors, calculated from the Force and Couple Vectors point towards the line of action shown in yellow. Notice that the larger Couple vector of Point 2 also has a larger radius or distance to the axis. This is because the magnitude of the determinate vector is equal to the distance from the axis. Another thing to note is that based on the Point being used to find the line of action, the determinate vector from that point will intersect the line of action axis at a different Z coordinate shown in light blue. This is ok for now because as the thread describes the aerodynamic forces acting along this line of action and therefore all points along the axis are valid.



Based on this information we have now found the Line of Action to describe how the aerodynamic forces are acting on the car. This will give us the X and Y coordinates for this line based on the reference frame or plane intersecting the origin. If we base these X,Y coordinates off of a plane passing through the CoG, then we can also find the height coordinate and we will have a theoretical and instantaneous “CoP” for that singular reference plane. However we must be careful how we convey this information as a singular Z coordinate because it is all subjective of the reference plane.

To show this, the next picture depicts the original Force and Couple vectors located at the CoG and the determinate vector pointing towards the line of action. The vector lies along the blue plane with the light blue dots showing where the plane intersects the CoG and the line of action. When taking the measurement for a Z coordinate from this position, it can be seen that the point intersecting the line of action is below the Global X,Y,Z coordinate system plane at (0,0,0).
The orange dot shows where the line of action intersects the global X,Y,Z plane. It also shows the determinate vector and orange point along the Z axis intersection. Also, it can be seen that the line of action is angled towards the Z axis and at some point in space will eventually intersect it. This means that based the Point used along the Z axis to determine the line of action location will not only change the Z value, but also the distances away from the axis in X and Y. This is further displayed through the distance between the Blue dots being smaller then the distance between the Orange. This fact further supports the original theory that there is no center of pressure, only a line of action around which all forces act.



To help reinforce the whole topic, the thread also shows this Force screw from the top of the line of action shown in the next picture. Since we are looking from above the line of action is a dot and parallel to all Red Force Vectors. It is now easier to see how the couple vectors are swirling around a central axis.



Yet another good example is the last picture in the thread which shows the Force screw but from an easier perspective where the Force and Couple Vectors are collinear on the Z axis and at the exact global center. Since the Blue couple vector is pointing in the positive Z direction, all other Couple vectors have a vertical component. The added light blue line is the vertical component of these couple vectors while the green lines are the circumferential component.



I hope someone on has come across this theory before and is able to help confirm its validity and possibly shed some more light on the correct workings. I have only this one thread as a reference so any help would be much appreciated!

Another example that may help to explain this theory is if we simplify the result to a 2D example, which would only be valid if for driving in a straight line with 0 Yaw moment.

The picture below shows the following - not drawn to scale and exaggerated values:

CoG

• A red Point CoG
• Light red CoG height values based on the specific X position along the car
• An average “line of action” for the CoG

Aerodynamic LoA

• Two Line of Action examples, LoA1 & LoA2, based on hypothetical aerodynamic force values for Drag Fx -Purple, and Downforce Fz - yellow, acting on the car and the resultant force Fr- dark green creating the Line of action.
• The intersection of LoA1 & LoA2 on the CoG line of action = Green dot with Red outline
• The intersection of the determinate Moment vector perpendicular to the Fr vectors calculated at the CoG shown as an orange dot with green outline, which lies on LoA.

These results are shown even better with a zoomed in picture of the CoG in the picture below. As we can see, if a force and moment calculation is taken at the CoG, this will result in 3 Force values in (X,Y,Z) and three moment values in (X,Y,Z). For this 2D example the Y values are ignored.

The orange line below is the resultant moment vector - Mr composed of the Moment vectors - Mx and - Mz outputs obtained in the forces and moments result control. Their positive or negative values are only used to show direction away from our 0,0 origin at the CoG.

It can also be seen, as described in the force screw theory, the Resultant Moment vector - Mr (orange) is perpendicular to the Resultant Force Vector Fr (dark Green). When we are reading the Moment Vector - Mx output from the result control with the origin at the CoG, it can be falsely perceived as the distance along the X axis where the “Center of Pressure lies”. This is shown as the white dot for both LoA examples. In reality, this X location along the CoG plane should be where the line of action intersects, shown as the green dot with red outline. With a hypothetical measurement, it can be understood that the X location actually lies further away for LoA 1 at 5mm and LoA 2 at 12mm. This difference doesn't seem like much but with increasing moment distances, the pitch, roll, or yaw moment forces significantly increase.



Again, any insight to my reasoning would be greatly appreciated!!

[ds4719]

Anyone have some input to this theory. Any help would be appreciated

[LuckyTran]

What is your query? These are concepts we usually teach immediately after introducing force couples in mechanics, very basic mechanics stuff.

Any force acting at a point (we can call this the CoP) at some distance from a center (we'll call this the CoG) can be replaced by a force (now acting from anywhere you like) and a force-couple. This theorem is so prevalent, I'm not sure it can even be attributed to any name. So yes, you can always choose to take an arbitrary field of forces and turn it into a srewdriver acting somewhere. You could also have chosen to put this screwdriver at the CoP; and if you had done this, it wouldn't have had any twist to it.

Btw, the CoG also needs to be calculated by the interaction of a mass gravity field and the gravitational field. That is, the CoG is also a screwdriver.

Originally Posted by ds4719

While this is useful in telling us overall stability, it does not give us information on the forces at the locations that really matter, which is directly at the contact patches of the tires.

And??? You are doing overall stability analysis. If you wanted to know what's happening at the contact patches, that's not overall stability. If you wanted to get that information, don't consider the system as a single rigid body but a connected system of smaller bodies (i.e. multiple components). And you still can use either method.

[sbaffini]

Too long to actually grasp from the text what you're actually asking, also considering that wikipedia has this well sorted out in few lines here:

https://en.wikipedia.org/wiki/Resultant_force

Maybe reformulate the thing more succintly or wait for someone in the mood

[ds4719]

LuckyTran thanks for your response. I havent gone that deep into mechanics so this topic was new for me. I have read a decent amount about aerodynamics and the CoP but NOT ONCE have i seen it related to this force-couple principle.

I made this post for another forum and copied it here because I felt it was necessary to understand this better with help from the experienced members here and for anyone else that has never heard of this relation.

So yes, you can always choose to take an arbitrary field of forces and turn it into a srewdriver acting somewhere. You could also have chosen to put this screwdriver at the CoP; and if you had done this, it wouldn't have had any twist to it.

This was the main point i wanted to understand. I can take a forces and moments measurement at any point in the simulation. I chose the CoG. The moment about this point should point towards this "Line of Action" and along this line there is no twist, as no moment just a pure force, got it. This concept was not to difficult to grasp.

And??? You are doing overall stability analysis. If you wanted to know what's happening at the contact patches, that's not overall stability. If you wanted to get that information, don't consider the system as a single rigid body but a connected system of smaller bodies (i.e. multiple components)

I agree that so far I am doing overall stability, weather measured from a point, the CoP, or forces on each axle. However I wanted to go a bit deeper into the relation to vehicle dynamics by calculating the force and force-couple at each contact patch. This is because front/rear tells us stability, but vertical forces at each tire can tell us how much extra travel the wheel has moved due to this force --> how much extra camber change has occurred --> how this effects slip angle --> leading to how much lateral force can be achieved.

The correct methodology and application of this concept is where i was looking for guidance.

sbaffini - i know its long and sorry for that. But thank you for the Wikipedia link, i also had trouble finding info on "force screw" probably because its not called that

-Dan

[LuckyTran]

You have to look in engineering mechanics bodies of text to find force couples, force wrenches, and force screws. Most formally it is called a force couple, but wrenches and screws are often mentioned for pedagogical purposes. In mathematics, these are just moments.

The force-couple does not provide any new information that you couldn't have known by knowing the force at the CoP and the CoG.

The principle of moving forces around in order to do any type of analysis goes back to free body diagrams. Knowing the CoP, the amount of the force, and the lever arm distance to the CoG, you can very quickly find the moment (the couple). Now you know the couple and can find the moment about any other point (such as the tires). You can then draw a free body diagram of your part, the tires, and expose the internal forces and very quickly find the "vertical forces" there.


These are all concepts we teach in introductory statics courses, which is usually the first true engineering course (that isn't a basic science like physics, calculus) to engineering students. E.g. here is a bridge with a load on it. Find the resultant forces on the two supports.

[Ship Designer]

Thanks @ds4719 for your exhaustive post! You shed light on this topic and mention some key points that are essential to understanding this. I've seen many posts about people asking how to find the Centre of Pressure and most of them sadly remain unresolved.

Some points that I believe contribute to cause confusion about this topic:

1. There are force systems that cannot be further reduced than to a resultant force and a moment that are not perpendicular to each other, the so-called wrench that you mentioned. They rather reduce to a resultant force and two resultant moments, one of which is perpendicular to the force while the other is parallel to it. The perpendicular moment can also be zero in some cases. This in a way contradicts the common definition that a moment has to be perpendicular to the force that generates it. Out of about 10 mechanics textbooks I've consulted, only one vaguely touches this concept. This video shows how the resultant moment can be decomposed into these two components, if they are present. So knowing that a resultant moment can be parallel to a resultant force is crucial, because then no matter where the line of action of the resultant force passes through, that moment won't vanish.
2. Most definitions I saw for Centre of Pressure, mention that it is a point about which all moments become zero. However, these definitions fail to mention, like you on the other hand demonstrated, that there are force systems for which no such point exists, i.e. where there will still be a moment that cannot become zero. An axis can be found where the resultant force passes through where its perpendicular moment vanishes, but the parallel one still remains. Not knowing this makes people wonder why they can't seem to find the CofP, even by iteratively guessing and approximating the chosen Centre of Rotation, where at least one moment component never becomes zero.

Thanks also for these:

Tx(P) = Tx(0) + (-Fz*Py + Fy*Pz)
Ty(P) = Ty(0) + (+Fz*Px - Fx*Pz)
Tz(P) = Tz(0) + (+Fx*Py - Fy*Px)

Hopefully they can be useful for obtaining moments by translating the CofR without having to restart the CFD Software. I'll try them out.

Cheers, Claudio

[LuckyTran]

This is what we call fake news.

All that plus the youtube video is literally the first two equations shown in the Wikipedia for https://en.wikipedia.org/wiki/Resultant_force

No one has lied to you about calculating resultant forces and resultant moments.

[Ship Designer]

Originally Posted by LuckyTran

This is what we call fake news.

All that plus the youtube video is literally the first two equations shown in the Wikipedia for https://en.wikipedia.org/wiki/Resultant_force

No one has lied to you about calculating resultant forces and resultant moments.

I must have missed the Wikipedia article, thanks for that.

Where exactly is my post fake Sir, if I might ask?

[LuckyTran]

You are claiming that that I cannot compute a resultant force and moment for systems of forces yet present equations showing exactly how to do that. The screwdriver is a moment (and also a force couple). Depending on which area in the world you grew up in, you'll call it by one of these terms or other terms. It doesn't mean that it's not the exact same thing.

[Ship Designer]

You are claiming that that I cannot compute a resultant force and moment for systems of forces yet present equations showing exactly how to do that.

I believe there's a misunderstanding. I stated that some force systems cannot be reduced further than to a wrench, i.e. a force and a moment parallel to it. I assume this is what you mean by screwdriver but please correct me if that's not the case.

[LuckyTran]

I think you are confusing the center of pressure with the center of rotation. They are not the same thing.

Moments are always with respect to their moment centers. The moment center can be arbitrarily placed anywhere. You shouldn't ever talk about a moment without specifying what is the momentum center. A lot of people assume that this moment center is your center of "insert the name of your favorite animal here" when it can be anything.

[Ship Designer]

I believe I'm aware of the distinction between the two. It is my understanding that if the Centre of Rotation is placed at the Centre of Pressure, then the resultant moment component perpendicular to the resultant force becomes zero. The Centre of Gravity can very well be somewhere else.

[LuckyTran]

If the center of rotation and pressure coincide to the same location yes. But you don't get to put the center of rotation anywhere. The body or body collection moves according to the forces acting on it. What you choose is the moment center. The resultant force always has a line of action which is whatever it is. Moments are calculated about arbitrary moment centers which may or may not be on this line of action. Notice I've only used the word moment. There are an uncountable number moments than be calculated for any force system. They can be calculated w.r.t. to the center of pressure, w.r.t. the center of rotation, w.r.t. to the end of the universe.

What you are referring to is torque, which is a very specific moment. The torque vector is the cross product of the force vector and the lever-arm vector. Maybe you didn't know that it was a cross product and thought that it was simply the product of the two magnitudes and left out the sin(theta) part, then I can see how it might be confusing. Otherwise, it's in the definition and the first equation that appears in the wikipedia article.


You don't place the center of rotation or the center of pressure anywhere. They are whereever physics says they will be. You can pick and choose the moment center though.

[Ship Designer]

I wasn't aware of the distinction used in the US between torque and moment. I'm only talking about moments. Please allow me to clarify. The Centre of Rotation I was referring to, is an arbitrary point I have to define in the CFD software I use, about which the resultant moment is calculated for one or more boundary patches that I select. I've derived a formula to accurately calculate the Centre of Pressure only from the resultant force and moment. The expectation was for the case I'm working on, as a means of verifying this, that if I place that point at the Centre of Pressure calculated by me (the software doesn't calculate it), the software would output a zero (or very small) resultant moment for the patch in question. But it doesn't, because the force system has a resultant moment component parallel to the force. At first I didn't know that and suspected something might be wring with the software, now I know it's not. It works well though e.g. for a 2D Airfoil where the force system is planar. What I'm working on however is more complex than that. So that's what I meant by Centre of Rotation and I should have been more clear about this earlier. My apologies.

You're right of course that the resultant forces and moments are what they are. But then it's my job to find out where they act and to make sure that they act at places that are appropriate for whatever it is that I'm designing.

If I go the long way and extract data from post processing, I can calculate the forces via pressure, surface areas and normals and the resultant moment by forming the cross product and get very similar results to what the software outputs. However, the resultant moment I'm left with when evaluating the moments around the calculated Centre of Pressure are in my opinion still far too large and so I'm not sure if that point I've calculated is really correct. That's what I'm still investigating.