# CG Center of Gravity

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#### compass1

##### Member
Just wanted to know if there is a formula for determining the cg of any rocket.
In rc aircraft the starting point is around 33% of the wing width, measuring from the leading edge.

#### GrossApproximator

##### Well-Known Member
Rocksim. Other than a computer program like that, I don't know of a formula--but then again, I'm no expert.

##### Well-Known Member
Just wanted to know if there is a formula for determining the cg of any rocket.
In rc aircraft the starting point is around 33% of the wing width, measuring from the leading edge.
I believe you mean the center of pressure, as the center of gravity is just finding the balence point of the rocket, this can be done by balencing the rocket on the ruler, the cente of pressure can be roughly estemated using the cardboard cutout method, which is balencing a cardboard template of your rocket on a ruler and finding that point, it gives you a very conservitve result, and, usually resulting in an over stable rocket. The barrowmans stability equations is the last method, but it assumes the drag your body tube and that you have 3 or more fins, finally there is the extended barrowmans equation which takes into account the body tube and less than 3 fins probobilty

#### TheAviator

##### Well-Known Member
I presume you mean where you want the CG to be. Ideally, it should be one caliber (largest airframe diameter) ahead of your center of pressure (CP), which is similar to the neutral point (NP) in model airplanes.

The simplest (and most conservative) way to calculate the CP is the Cardboard Cutout method. Simply cut out a silhouette of your rocket in a piece of cardboard and balance it on a knife blade. The place where it balances is the CP at a 90 degree angle of attack. Because this is the farthest forward the CP can ever be, your rocket will always be stable if your CG is further forward than the CP.

##### Well-Known Member
I presume you mean where you want the CG to be. Ideally, it should be one caliber (largest airframe diameter) ahead of your center of pressure (CP), which is similar to the neutral point (NP) in model airplanes.

The simplest (and most conservative) way to calculate the CP is the Cardboard Cutout method. Simply cut out a silhouette of your rocket in a piece of cardboard and balance it on a knife blade. The place where it balances is the CP at a 90 degree angle of attack. Because this is the farthest forward the CP can ever be, your rocket will always be stable if your CG is further forward than the CP.
Beat you but there is somthing extremely strange, I got .71 stabitlty with extended barrowmans but got 3.41 on the cardboard cutout method

#### RandyT0001

##### Well-Known Member
Just wanted to know if there is a formula for determining the cg of any rocket.
In rc aircraft the starting point is around 33% of the wing width, measuring from the leading edge.
http://www.spacemodeling.org/JimZ/manuals/tir-33.pdf
That old tech report has a method of calculating the center of gravity (cg) and the center of pressure (cp) of a rocket. Read it over a few times following along with the math for the examples and you soon learn how to calculate cg and cp. It starts the cg calculations on p 32. Don't forget to include the mass of the engine when determining cg (an oft overlooked component).

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#### JRThro

##### Well-Known Member
TRF Supporter
Just wanted to know if there is a formula for determining the cg of any rocket.
In rc aircraft the starting point is around 33% of the wing width, measuring from the leading edge.
The absolute best way to determine the CG of a rocket is to install the heaviest motor you plan to use in it, and then balance the rocket on a straightedge.

That'll show you where the CG really, truly, is.

#### georgegassaway

Going by the title of this thread:

The CG is usually located in the middle of the Center of Gravity....

- George Gassaway

#### RandyT0001

##### Well-Known Member
IIRC these calculation methods (cg & cp) were included in some older versions of the Handbook of Model Rocketry (3rd ed 1970 or 4th ed 1976 which I had but lost in a move while out of rocketry) but when I purchased my latest copy (6th ed 1994) the cg calculations had been deleted. Are they in the latest 7th edition?

#### BEC

##### Well-Known Member
And in the model airplane world 33% of the chord back from the leading edge is an approximation of where it should be, not necessarily where it is. Actually that's a bit aft for many airplanes....

John's answer is the way to go to find out where it IS for a given rocket. Some of the discussion about simulators and cardboard cutouts and cranking through the Barrowman equations sheds light on where it should be.

#### cjl

##### Well-Known Member
You gotta love tHe archive.... One of these days I'll have to check it out
If you feel like starting a thread, definitely search the archive first. It contains an immense amount of useful information, plus over 8000 posts of my useless blathering

#### compass1

##### Member
And in the model airplane world 33% of the chord back from the leading edge is an approximation of where it should be, not necessarily where it is. Actually that's a bit aft for many airplanes....

John's answer is the way to go to find out where it IS for a given rocket. Some of the discussion about simulators and cardboard cutouts and cranking through the Barrowman equations sheds light on where it should be.
I always like to set my cg on planes at 33% it makes it more fun to fly when they are touchy.

#### stantonjtroy

##### Well-Known Member
There have been many links and directions, ALL OF THEM GOOD. However, the simple and direct answer to your question is; The CG (Center of Gravity) is, literly where it is. It is the balance point of the rocket, or if you prefer, the point where all mass force vectors equal zero. There is a means of calculating it however I think you would like to know where it SHOULD be. This is where the CP (Center of Preasure) calculations that have been recomended come in to play. The reference of 1 "Caliber" means to balance the rocket 1 "Body Tube Diameter" ahaed of the CP. A 1-2.5 Caliber range is workable. It is the CP/CG relationship that is important to stability.

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#### powderburner

##### Well-Known Member
And adding on a little bit to the end of Troy's post, c.g. location is NOT found through the use of Barrowman eqns or cardboard cutouts. Those tools are for estimation of c.p., or center of pressure.

Some of the discussion about simulators and cardboard cutouts and cranking through the Barrowman equations sheds light on where it should be.

#### light

##### New Member
In the Name Of God..

Center of gravity of rocket equals the sum of the weight times the distance of the components divided by the rocket weight.

As a rocket flies through the air, it both translates and rotates. The rotation occurs about a point called the center of gravity. The center of gravity is the average location of the weight of the rocket. The mass and weight are distributed throughout the rocket, and for some problems, it is important to know the distribution. But for rocket trajectory and maneuvering, we need to be concerned with only the total weight and the location of the center of gravity.

How would you determine the location of the center of gravity?

Calculating cg

A model rocket is a combination of many parts; the nose cone, payload, recovery system, body tube, engine, and fins. Each part has a weight associated with it which you can estimate, or calculate, using Newton's weight equation:

w = m * g

where w is the weight, m is the mass, and g is the gravitational constant which is 32.2 ft/square sec in English units and 9.8 meters/square sec in metric units on the surface of the Earth. On the Moon and Mars, the gravitational constant and the resulting weight is less than on Earth. To determine the center of gravity cg, we choose a reference location, or reference line. The cg is determined relative to this reference location. The total weight of the model rocket is simply the sum of all the individual weights of the components. Since the center of gravity is an average location of the weight, we can say that the weight of the rocket W times the location cg of the center of gravity is equal to the sum of the weight w of each component times the distance d of that component from the reference location:

W * cg = [w * d](nose) + [w * d](recovery) + [w * d](engine) + ...

The center of gravity is the mass-weighted average of the component locations.

Components' Location

On the slide, we show the weight and distance of the nose cone from the reference line. A similar distance can be determined for each component relative to the reference line. How do we determine the distance d? Using the nose cone as an example, the "distance" of the nose dn is the distance of the cg of the nose relative to the reference line. So we have to be able to calculate or determine the cg of the nose cone and each of the other rocket components. For some simple shapes, finding the cg, or average location of the weight, is quite simple. For example, when viewed perpendicular to the axis, the body tube is rectangular. The cg is on the axis, halfway between the end planes. For other shapes, like the nose cone, determining the cg of the component is not so simple. There is a technique for determining the cg of any general shape, and the details of this technique is given on another page.

Determining cg Mechanically

For a small model rocket, there is a simple mechanical way to determine the cg for each component or for the entire rocket:

1. For simple geometries we just balance the component or the entire rocket using a string or an edge. The point at which the component or rocket is balanced is the center of gravity. This is just like balancing a pencil on your finger! Obviously, we could not use this procedure for a large rocket like the Space Shuttle, but it works quite well for a model.
2. Another, more complicated way, is to hang the model from some point, for example, the corner of a fin, and drop a weighted string from the same point. Draw a line on the rocket along the string. Repeat the procedure from another point on the rocket, the nose, for example. You now have two lines drawn on the rocket. The cg is the point where the lines intersect. This procedure works well for irregularly shaped objects that are hard to balance. The problem with this procedure is that the cg can fall outside the body for complex geometries.

Just wanted to know if there is a formula for determining the cg of any rocket.
In rc aircraft the starting point is around 33% of the wing width, measuring from the leading edge.
:blush:

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#### light

##### New Member
Computer drawing of a variety of objects showing the center of gravity - CG. CG = sum of component weight times component distance divided by total weight.

The center of gravity is a geometric property of any object. The center of gravity is the average location of the weight of an object. We can completely describe the motion of any object through space in terms of the translation of the center of gravity of the object from one place to another, and the rotation of the object about its center of gravity if it is free to rotate. In flight, rockets rotate about their centers of gravity.

Determining the center of gravity is very important for any flying object. How do engineers determine the location of the center of gravity for a rocket which they are designing?

In general, determining the center of gravity (cg) is a complicated procedure because the mass (and weight) may not be uniformly distributed throughout the object. The general case requires the use of calculus which we will discuss at the bottom of this page. If the mass is uniformly distributed, the problem is greatly simplified. If the object has a line (or plane) of symmetry, the cg lies on the line of symmetry. For a solid block of uniform material, the center of gravity is simply at the average location of the physical dimensions. For a rectangular block, 50 X 20 X 10, the center of gravity is at the point (25,10, 5) . For a triangle of height h, the cg is at h/3, and for a semi-circle of radius r, the cg is at (4*r/(3*pi)) where pi is ratio of the circumference of the circle to the diameter. There are tables of the location of the center of gravity for many simple shapes in math and science books. The tables were generated by using the equation from calculus shown on the slide.

For a general shaped object, there is a simple mechanical way to determine the center of gravity:

1. If we just balance the object using a string or an edge, the point at which the object is balanced is the center of gravity. (Just like balancing a pencil on your finger!)
2. Another, more complicated way, is a two step method shown on the slide. In Step 1, you hang the object from any point and you drop a weighted string from the same point. Draw a line on the object along the string. For Step 2, repeat the procedure from another point on the object You now have two lines drawn on the object which intersect. The center of gravity is the point where the lines intersect. This procedure works well for irregularly shaped objects that are hard to balance.

If the mass of the object is not uniformly distributed, we must use calculus to determine center of gravity. We will use the symbol S dw to denote the integration of a continuous function with respect to weight. Then the center of gravity can be determined from:

cg * W = S x dw

where x is the distance from a reference line, dw is an increment of weight, and W is the total weight of the object. To evaluate the right side, we have to determine how the weight varies geometrically. From the weight equation, we know that:

w = m * g

where m is the mass of the object, and g is the gravitational constant. In turn, the mass m of any object is equal to the density, rho, of the object times the volume, V:

m = rho * V

We can combine the last two equations:

w = g * rho * V

then

dw = g * rho * dV

dw = g * rho(x,y,z) * dx dy dz

If we have a functional form for the mass distribution, we can solve the equation for the center of gravity:

cg * W = g * SSS x * rho(x,y,z) dx dy dz

where SSS indicates a triple integral over dx. dy. and dz. If we don't know the functional form of the mass distribution, we can numerically integrate the equation using a spreadsheet. Divide the distance into a number of small volume segments and determining the average value of the weight/volume (density times gravity) over that small segment. Taking the sum of the average value of the weight/volume times the distance times the volume segment divided by the weight will produce the center of gravity.

#### BEC

##### Well-Known Member
And adding on a little bit to the end of Troy's post, c.g. location is NOT found through the use of Barrowman eqns or cardboard cutouts. Those tools are for estimation of c.p., or center of pressure.

Fair enough... I should've been more clear and was thinking in model airplane terms where one moves stuff around to get the balance point in the right place relative to the aerodynamic configuration of the airplane rather than in model rocketry terms where often the aerodynamic configuration may be in play as much as the mass distribution within it. But either way, correcting a tail heavy (or insufficient stability margin) situation can be done the same way - add nose weight. Using the CP calculations just suggest where the CG should be by using CP as a necessary reference point, which is what I was thinking.

#### light

##### New Member
Fair enough... I should've been more clear and was thinking in model airplane terms where one moves stuff around to get the balance point in the right place relative to the aerodynamic configuration of the airplane rather than in model rocketry terms where often the aerodynamic configuration may be in play as much as the mass distribution within it. But either way, correcting a tail heavy (or insufficient stability margin) situation can be done the same way - add nose weight. Using the CP calculations just suggest where the CG should be by using CP as a necessary reference point, which is what I was thinking.
Of course..

*For a model rocket, there are some simplifying assumptions that we can use to make this task much easier. Model rockets are fairly symmetric about the axis of the rocket. This allows us to reduce the full three dimensional problem to a simple, two dimensional cut through the axis of the rocket. For model rockets, the magnitude of the pressure variation is quite small. If we assume that the pressure is nearly constant, finding the average location of the pressure times the area distribution reduces to finding just the average location of the projected area distribution.

..To resolve some of design problems, engineers prefer to characterize the forces on an object by the aerodynamic force, described above, coupled with an aerodynamic moment to account for the torque. It was found both experimentally and analytically that, if the aerodynamic force is applied at a location 1/4 of the length back from the leading edge on most low speed objects, the magnitude of the aerodynamic moment remains nearly constant with angle of attack. Engineers call the location where the aerodynamic moment remains constant the aerodynamic center. Using the aerodynamic center as the location where the aerodynamic force is applied eliminates the problem of the movement of the center of pressure with angle of attack in aerodynamic analysis. (For supersonic airfoils, the aerodynamic center is nearer the 1/2 chord location.)

When computing the stability of a rocket, we usually apply the aerodynamic forces at the aerodynamic center of airfoils and compute the center of pressure of the vehicle as an area-weighted average of the centers of the components..

http://exploration.grc.nasa.gov/education/rocket/rktcp.html
http://exploration.grc.nasa.gov/education/rocket/cp.html