Center of Mass(CoM) by the pic method.

[Aussie_Paul]

I have been wanting to get around to this for a while. I think that it is important that if someone is interested enough to want to calculate the CoM of their gyro, the method should be available.
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Finding a gyro’s Center of Mass by the pic method. 25/04/04

A number, minimum of two, pictures are required of the gyro suspended or balanced from at least two different points.
During the hang test a picture is taken exactly from the side. The more rope above the teeter bolt the better, as this makes it easier to draw an accurate line. Pic (1).

With the rotors and pilot on board the machine is balanced on blocks, 20liter drums or something of a similar height and strong enough as to not collapse. This can require a person at the front and back of the gyro. While the gyro is balanced another picture is taken again exactly from the side. Pic (2).

Print picture (1) and draw a line down the supporting rope and continue the line down to at least the level of the keel.

Print Picture (2) and draw a vertical line from the center of the axle at least up past the propeller line.

As you will see in my pictures I have a corrugated iron wall to use as a vertical reference point. Just make sure the builder built it straight. The machine has to be at right angles to the wall.

Now we have to superimpose the line from picture (1) onto picture (2) or vica versa.

Where the two lines cross is the Center of Mass (CoM) of that gyro at that weight.

Extend the propeller thrust line forward past the CoM point.

The distance between the CoM and the extended propeller thrust line is the amount of offset between the two. The distance can be scaled off the picture quite easily.

This is the figure that is bantered around the traps in relation to pitch stability and Power Push Over (PPO) forces.

This article is written with the sole aim of helping people, if they are so inclined, to measure the offset.

Paul Bruty.

[Aussie_Paul]

This is the last pic with the lines superimposed onto the one pic.

Aussie Paul.

[Aussie_Paul]

Pic number 1 Aussie Paul.

[Aussie_Paul]

pic number 2. Aussie Paul.

[Aussie_Paul]

Gee, I would have thought that this was a very important piece of information that everyone would want to know about their gyroplane.

Aussie Paul.

[quadrirotor]

super, really!
now, how do you determine the center of drag?

[Aussie_Paul]

Andre, that is beyond me at this time. At the speeds that a gyro normally flys, up to 60 mph, the CoM and the thrust line offset are the most important. As we go faster, and there are now machines capable of greater speeds, then the CoDrag becomes more important. To have the Thrust line, the CoM, and the CoD within a couple of inches of each other, with an effective h/stab, is the ultimate.

If the average person building an open frame single seater gets the Thrust line and the CoM with a couple of inches, has some sort of a h/stab, and receives basic training, they are guaranteed of success.

Aussie Paul.

[BenMullett]

You are probably right, Paul. And a misaligned aerodynamic force from a miscalculated centroid of drag is more amenable to a balancing aerodynamic force (from a stab, eg) than, say, a thrust misalignment.

I must say that I agree with Greg that it matters little how we achieve our stability - so long as we have a repeatable method (or methods) that work(s) and whose function is understood.

All the best, Ben

[rehler]

Don,

The rotor blades (on or off) do not change the CG "line", since the line is going through the center of the rotor blade's CG. Hanging from the teeter bolt does not show where the actual CG is located, only the "line" that goes through it.

The blades do need to be installed for the other balance.

[Hognose]

Don said: "The fore-aft CG change isn't all that different, only a few degrees difference."

Don, your illustrations show 0.7 degrees different. Am I missing something?

The principle behind Paul's rough-and-ready CoM calculation is:

- if you hang the machine from the teeter bolt, a vertical line that intersects the teeter bolt must run through the centre of mass.

- and if you balance the gyro on its two mainwheels (like a unicycle!), a vertical line through the center of the mainwheels must run through the centre of mass.

The point where the two lines intersect must be the (2-dimensional) centre of mass. This lends itself to a number of relatively simple tests to establish a centre of mass range. How does the centre move when fuel is burned off? What if the pilot weighs 220lb (100kg)? What if the pilot is a 110 lb (50kg) girl?

You could determine what error there is in this calculation by adding a third balance, on the nosewheel with several supporting the gyro but not bearing any of its weight (obviously this is more theoretical than practical). In this case a vertical plumb line through the nosewheel hiub must run through the centre of mass and intersect the other lines. Assuming the lines don't all perfectly meet at the same point, the smallest circle that could encompass the triangle at the intersection of the three lines is your margin of error.

cheers

-=K=-

[Al_Hammer]

Don,
I think maybe you're thinking that the cg moves horizontally when the rotor is added. In the drawing below, you can see that the cg moves along the line that was determined in the hang test. This movement has a horizontal component when you tilt the gyro, but the cg is still on the line. The exact point is found by combining the results of the hang test and the balance test.

hangtest_2 shows another method you might like.

[rehler]

Al,

I respectfully expel the notion that hanging a gyro from the teeter bolt will place the CORRECT CG directly under the point of hang. It might be close, but it's not going to be exact.


Don,

Having the rotor on or off does not affect the line from the teeter bolt to the CG when hanging the gyro from the teeter bolt. The CG must be EXACTLY on the line, not close to it but exactly on it.

The rotor blades are balanced on the teeter bolt so their weight does not affect the vertical CG line while hanging. The blade CG is also exactly on the line.

Ken

[Vance]

Hi Ken and Don, I am getting confused. I would think that for calculating the vertical cg for centerline thrust it would be the cg with the rotor blades on. When calculating the cg to make sure the rotor thrust vector remains behind the cg it would be with the rotor off. When you throw in aerodynamic drag and the forces in foward flight I get realy lost. I would be grateful for some knoledge to unconfuse me. Thank You, Vance

[Al_Hammer]

Vance, I think the cg with rotor blades on is what you want for all calculations of moments about the cg. Even though the rotor is lifting its own weight, etc, it is still part of the total mass of the aircraft.

The primary forces in flight to be concerned with are prop thrust, rotor thrust( which is the rotor lift and drag represented as a vector) and drag of the fuselage.
Weight is also important, but it always acts through the cg. The other forces may not pass thorugh the cg, so the moments need to be calculated for each force.

I hope I'm not adding to the confusion. :)

[ymmv]

I respectfully expel the notion that hanging a gyro from the teeter bolt will place the CORRECT CG directly under the point of hang. It might be close, but it's not going to be exact.

Don, anything you hang from a single point will align itself such that the CG is directly below the hang point -- be it a gyro or a teacup. A vertical line extending through the hang point always includes the CG. Hang the object from a different point and you'll get another vertical line along which the CG must lie. The two lines will intersect, and their intersection is the CG point.

The reason that the rotor doesn't affect the teeter bolt hang test is that it is symetrically disposed on either side of the hang point. It balances, so it doesn't change the hang angle of the gyro. Hang the gyro from any other point and the weight of the rotor will matter; but from the teeter bolt, no.

A more accurate hang test would be to hang the gyro from some point on the mast or cheek plates, with the rotor blades on.

Not more accurate, it just gives another line along which the CG must lie. The intersection of that line with the teeter-bolt-hang line is the CG.

[PW_Plack]

Don,

The line created by balancing on the wheels will change from blades-on to blades-off, but not the line created by hanging from the teeter bolt. The influence of the blades on CG will be correctly reflected by the change in the main gear balance point, and the resulting change in the point of intersection of the two lines.

Same thing would hapen if you took the wheels off the main gear for the balancing test, and just balanced the gyro on the axle. This WOULD change the angle of the hang test, but not the angle of the balance test.

[Vance]

Hi Al, Yes, I am more confused, but on a higher level. I want to make sure that I understand what you are saying. As I understand it the rotor thrust vector is at ninety degrees to the rotor and it needs to pass thru of behind the center of gravity in all modes of flight. The center of gravity is raised and moved rearward by the weight of the rotors and rotor head compared to the the gyro without the rotors or rotor head.

I can see why you would want to always pull it's tail rather than it's nose. How much angle do I want to imagine for the rotor thrust vector? I realize that this has been answered many times before but in my opinion the answers haven't been consistant or have been using a baseline that I don't have. I would be grateful for any help and especialy any explaination of the differing opinions.
Thank You, Vance

[donshoebridge]

Let me start by saying that I completely understand that the CG will align itself to a point directly under the point of hang. I wasn’t in disagreement with that. What I WAS in disagreement with is that the CG moves along the hang line. Reason being, the rotor blade weight is not factored into the total gyro/man mass because the entire weight of the rotor blades is hanging from the same point that the rest of the gyro is, effectively negating the rotor blades from the actual location of the CG. What I couldn’t logically grasp is the fact that the CG moves along (as Al Hammer pointed out to me) the hang line for some reason that I have yet to understand. I didn’t think it was possible for this to happen, but it does. But I digress…

I would like to sincerely apologize for my absolute stubbornness in believing what everyone has been saying with regard to determining your CG location via a hang test. For some reason, I wouldn’t let myself believe that it could be done this way because the CG does in fact drop several inches when the gyro is hung. I have confirmed this fact by running computer CG calculations on my Hornet design – one fully loaded, and one with the rotor blades suppressed from the model. I’ve attached a gif file of my findings.

The attached image is now the placemat for the Crow that I am now eating. I should know better than to go into someone else’s sand box and kick sand in everyone’s face. Sorry

To avoid confusing anyone further, I have self censored myself and deleted all but this post.

Don

[Hognose]

To avoid confusing anyone further, I have self censored myself and deleted all but this post.
Don

Don,

Nice picture ;). Nicer still is that you have all your data in CAD/CAM format so that you can explore these things. There's a lot to be said for putting high tech to work on our low tech gyroplanes. But it must have been a hell of a lot of work to get the data into the machine.

In a way I wish you had left the other posts -- because others will come, wrestle with the idea as you did, and mightn't it be easier for them if they can follow your reasoning? Ah, well, it's "runway behind us" now.

A while ago someone asked about drag forces on a rotorcraft airframe. Total drag is the sum of induced and parasitic drag. Induced drag is created by the lift of the rotors (and it's one of the primary things holding gyro speeds down). Parasitic drag comprises form drag and skin-friction drag. The most hazardous drag condition in a gyroplane would be one where the rotor is suddenly unloaded while the machine is at high speed (parasitic drag increases with speed). If the centre of pressure is below the centre of mass, you have an overturning moment created by the drag working against inertia alone (even with zero thrust!). This drag effect is additive to any overturning moment caused by a high thrustline.

Drag can be guesstimated based on the flat plate area of the craft, but it can also be calculated rather accurately based on speeds achieved in flight test, at a known thrust level.

cheers

-=K=-

[PW_Plack]

Kevin,

Would the dangers of getting center-of-drag below center-of-mass not place a limit on how high a designer raises the center-of-mass? In other words, could one not over-do raising the center of mass in pursuit of CLT?

[donshoebridge]

To all those that still don't get it like I now do, try to get your mind around this...

Regardless where you pick up the gyro, the CG will always swing to a point directly under the point of hang. Therefore, any weight that is either added or removed from the gyro, AS LONG AS IT IS DONE ON THE LINE BETWEEN THE HANG POINT AND CG, the CG will always be on the original line somewhere. I don't know if the same would be true if you add weight above the point of hang. But for all practical purposes, there would be no reason to do this.

When someone hangs a gyro from the teeter bolt, it doesn't matter if the rotor blades are on or not because the angle difference between the two CG locations is zero degrees. The only thing that changes is the load on the teeter bolt. DUH!

He's a question for everyone to ponder... Since a gyro flys by the rotors picking up the gyro from the teeter bolt, would the CG location be calculated at the lower position because the rotor blades are not part of the total weight of the gyro?

If the answer is yes, seeing how the CG would climb as the rotor blades were unloaded because of the added rotor blade weight, wouldn't this help the thrustline offset issue?

If both of these are true, I can see how a gyro may hane more of a tendency to PIO because the CG is moving up-and-down depending on the degree of load on the rotor blades.

[Al_Hammer]

Wow, I love it when somebody eats crow around here. Good thing it wasn't me this time.

He's a question for everyone to ponder... Since a gyro flys by the rotors picking up the gyro from the teeter bolt, would the CG location be calculated at the lower position because the rotor blades are not part of the total weight of the gyro?

That's what Vance was asking also. No, the cg as calculated by the hang and balance tests is what you use.
The rotor is producing lift/thrust which acts on the entire mass of the gyro plus rotor. Doesn't matter that the rotor itself is not hanging from the teeter bolt, it is still part of the total mass and contributes to the total moment of inertia.


seeing how the CG would climb as the rotor blades were unloaded because of the added rotor blade weight, wouldn't this help the thrustline offset issue?

When the rotor is unloaded, the weight changes, but the mass is unchanged. The cg does not move at all, only the rotor thrust.

[Al_Hammer]

There's one more thing, Don.

You show the rotor in the drawing at about 10 degrees flying angle. I measured from tip to tip. That's within a degree or so of what a typical rotor might cruise at. Because of flapping, the rotor is tilted back a couple of degrees from the plane of the rotorhead.
The RTV is therefore at 10 degrees from vertical, not at 7+ degrees as you have depicted it.
I think you might need to adjust the cg so that the hang angle is closer to the standard 12 degrees. The way it is now, if the rotor is at 10 degrees, the gyro would need to fly nose high to maintain the RTV behing the cg.

Just thinking out loud here, correct me if I'm wrong. :eek:

[Vance]

Thank You Al, That helps validate previous fuzzy understandings. I am still confused about all the confushion. It would seem to be a fairly simple concept.
Thank You,Vance

[Al_Hammer]

Vance,
simple concept? I don't think so. Gyros are a riddle wrapped in an enigma.
I guess thats part of the appeal.

[Vance]

Al, I was only speeking of where the rotor thrust vector should be. It wouldn't hold my interest if it was less complex and interactive than it is. I think that it is wonderful that so many different aproaches and compromises seem to fly reasonably well. Thank You, Vance

[donshoebridge]

Al,

I haven't even started on making the cheek plates because I figured that I'd have to make some small shanges to the CG location anyways, which will be the last thing I do.

Don

[GeneWeber]

Hi Al,

A question and a comment.

Because of flapping, the rotor is tilted back a couple of degrees from the plane of the rotorhead. The RTV is therefore at 10 degrees from vertical …
If the plane of the rotorhead is at 10 degrees, but because of flapping the blades are really at 12, is the RTV 10 or 12?


I think you might need to adjust the cg so that the hang angle is closer to the standard 12 degrees.
You mean adjust the cg shown on the drawing, right? In real life the cheek plates would be made to create the desired hang angle.

Best Regards,

Gene

[rehler]

Don,

Looking at your drawing I see you show the nose wheel up 1.4 degrees. It is best for takeoff to have it sitting on the ground (gyro level).

Also, it looks like your horizontal stabilizer is level. To have the CG move in front of the rotor thrust vector it needs to tilt nose down, to pull the tail down and swing the CG forward in front of the rotor thrust vector. Otherwise the CG will be exactly in line with the RTV.

You indicate a CG in front of the rotor thrust vector on the drawing. This will not happen when the gyro is not in flight. The CG will be exactly on the rotor thrust vector unless some outside force moves it (HS pushing down to swing it forward and up for example).

I doubt it is possible to accurately determine the location of the gyro’s CG by calculations without the gyro being built and then weighed and balanced. I tried it often but never was close when the gyro is finally built. Too many things change the balance (weight of instruments, wheels, misc. braces, tail feathers, etc.). However, for design purposes I recommend you use some “standards” such as the rotor head axis being 9 degrees tilted back. The flapping adding 2 degrees back, making the rotor thrust vector 11 degrees back. This changes depending on how fast you are flying, etc., but is a good start for design.

So find where you can best calculate the gyro CG and place the teeter bolt 11 degrees back from it. Then tilt the HS nose down about 2 degrees.

Hope this is of help.

[Vance]

Hi Ken, I love it when you share specific information that way. I am looking foward to visiting with you at the fly in next weekend. I hope you can find a little time. I enjoyed my last visit with you a lot. I should be rolling in thursday evening. Thank You, Vance

[rehler]

Vance,

I am very disappointed, but my gyro is still not ready to get back in the air, so it will not be at the Southwest Regional Fly-in. It will still be sitting in my garage. It should be a great event, but I'm not sure how many gyros if any will attend this year, as I have not heard for other gyro pilots planning to attend.

[donshoebridge]

Also, it looks like your horizontal stabilizer is level.


It's not.


I doubt it is possible to accurately determine the location of the gyro’s CG by calculations without the gyro being built and then weighed and balanced. I tried it often but never was close when the gyro is finally built.


What method/program were you using that didn't work for you?

I've been using 3D CAD for almost 15 years now, mainly AutoCAD, SolidWorks, and SolidEdge. I know what works and what doesn't. I've had plenty of time to figure out how best to make the software do the work for me. And a lot of that time has been spent figuring weights and balances on assemblies.


Too many things change the balance (weight of instruments, wheels, misc. braces, tail feathers, etc.).


Granted! This is why you make accurate 3D models through reverse engineering and weighing of the off-the-shelf components. SolidWorks is a $5000 3D engineering package, not some paint program that comes bundled with Windoze. And yes, that was ment to be a small slam to those that use a paint program or illustrator type program to do engineering, ie. Apple and Mac type products.

[rehler]

Don,

It's not the software that is the problem. Computers are "garbage in garbage out". The problem is the input information. It is practically impossible to weigh all parts of the gyro without having the parts to weigh.

There are just too many things that are not known until the gyro is built; things such as wires, instruments, bolts, nuts, washers, brakes, radio, antenna, bearings, trim, etc. Based on my experience (I did accurately weigh the major componemts), I believe you too will find a difference between the completed gyro and calculations.

[Al_Hammer]

If the plane of the rotorhead is at 10 degrees, but because of flapping the blades are really at 12, is the RTV 10 or 12?

It would be at 12. The Rotor thrust is perpendicular to the actual tip path plane.


You mean adjust the cg shown on the drawing, right? In real life the cheek plates would be made to create the desired hang angle.

The RTV should be shown to be perpendicular to the rotor tip path plane.
The cg is accurately determined from Don's calculations, so it shouldn't move.

[donshoebridge]

Major components?!?! If that's all you're going to use in the calculation, then obviously it's going to be wrong. The only thing I don't have in the total weight of my model is electrical, instruments, and seat belts. Nut, bolts, washers, bushings, EVERYTHING else is there, which Includes the epoxy and Kevlar. All of the items that I have in my model have real world numbers that can be used such as material densities, hex sizes, washer diameters, wall thickness of tube, even rudder cable weights. The information is there, and if not, it's not that difficult to either find or figure.

Unlike most of the other aircraft projects of the world, I'm trying to figure out as much as possible up front so I'm not making scrap later, which seems to be more the norm, rather than the exception. I don't plan on spending $15,000 to build something that flies. If I can help it, I'm going to make the parts right the first time, and not keep making them over or working them till they do fit. I'm not going to go through the crap that Paul Jr. on American Chopper goes through just to build a bike. If I can figure it out up front, then that's what going to happen.