Under Standing Centre of Gravity
In flight, both airplanes and rockets rotate about their centres of gravity. Determining the centre of gravity is very important for any flying object. So how do engineers determine the location of the centre of gravity for an aircraft which they are designing?
In general, determining the centre of gravity (CG) is a complicated procedure because the mass (and weight) may not be uniformly distributed throughout an object. Non-uniformly shaped objects require the use of mathematical calculus, which we will not discuss as it is outside the scope of the syllabus and has already been calculated for our individual aircraft structure.
However, 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 material that is uniformly shaped, the centre of gravity is simply at the average location of the physical dimensions. (E.g. a rectangular block, 60 X 30 X 20, the centre of gravity is at 30, 15, and 10).
Here are two simple mechanical methods for determining the centre of gravity:
If we balance an object using an edge, the point at which the object is balanced is the centre of gravity. (Just like balancing a ruler on your finger.)
A more complicated method to finding the centre of gravity of an irregular shaped object is to:
- Hang an object from any point and then drop a weighted string from the same point. Draw a line on the object along the string.
- Repeat the above procedure from another point on the object. You now have two lines drawn on the object which intersect. The centre of gravity is the point where the lines intersect. This procedure works well for irregularly shaped objects that are hard to balance.
Bell206L II Long Ranger, CG
A helicopter can be compared to a pendulum, the point of suspension being where the main rotor hub intersects the mast and the pendulum weight being the helicopter. If the pendulum weight is allowed to stop moving, it will come to rest directly below the suspension point.
The Empty Helicopter centre of gravity has already been calculated by professional technicians through adding or removing ballast weight.
The balance position for the Empty Helicopter is measured from a single permanent Datum (a reference point). In our case the datum is one inch or twenty five millimetres in front of the nose of the aircraft. The measured distance from the datum to each Fuselage Station (F.S.) is called an Arm.
Fig-1: Empty Helicopter Showing Arms from the Datum to each Fuselage Stations (F.S).
There are four specific fixed fuselage stations, which we can add or move weight to.
The names of these stations are:
- Row 1: 2 Forward Facing Crew Seats or Front Seats.
- Row 2: 2 Aft Facing Seats or Mid Seats.
- Row 3: 3 Forward Facing Aft Seats or Rear Seats.
- Locker: Baggage Compartment.
The arms (distances) for each fixed Fuselage Station are:
Note: Each arm can be found within the B206L Handbook on pages 30 and 31 in the Table of Moment title columns.
Calculating Bell206L II, CG
Adding or moving weighted items in the empty helicopter will move its centre of gravity. We will need to recalculate the new centre of gravity position. Then plot the CG on the Longitudinal Centre of Gravity Fuselage Station graph to ensure it remains within allowable limits. Any change in centre of gravity during flight must also remain within the graphed envelope.
For example: If the CG of a helicopter is aft of the hub-mast intersection (weighted aft of the mast), the helicopter will be tail-down in flight. The pilot can correct this condition by moving the cyclic control forward.
If the required movement of the cyclic is great enough, the pilot will have used all available control, thus limiting manoeuvrability and forward speed. A loss of manoeuvrability is unsafe and care must always be taken to keep the centre of gravity within operational limits.
Note: Longitudinal Centre of Gravity Fuselage Station Graph found within the B206L Handbook on page 6.
The Empty Helicopter arm and balance point (centre of gravity) is located 130.8 inches or 3322.3 millimetres from the datum and the helicopter empty weight is 2250 lb or 1020.6 kg.
Engine Oil is a fixed item that must be added to the helicopter with an arm of 205 inches or 5207 millimetres from the datum and weighing 13 lb or 5.9 kg. Note: These weights and arms can be found in the table on page 26 or 27 of the handbook.
The front Crew Seats must have a minimum of 170 lb (77 kg) or greater of ballast added to it for flight. A pilot or combined crew weight greater than 170 lb will suffice.
When adding weight to the B206L-1 Long Ranger II you need to find the new CG. To find the new CG, total all weights that have been added to the helicopter including the Helicopter Empty Weight (EW).
The product of the weight of item A and its Arm from the pivot point (CG) produces a counter clockwise torque moment and the product of the weight of item B and its Arm produces a clockwise torque moment.
When the clockwise torque moment is equal the counter clockwise torque moment, the helicopter is balanced. Any moment in front of the centre of gravity is considered a negative value, having a negative Moment Arm, while and torque behind the centre of gravity is considered to be a positive value with a positive Moment Arm.
Finding the balance point: If we wish to find the balance point (CG) for this system, we first must set up an arbitrary zero reference point (Datum). For this example, we will choose a zero reference point to be at the Datum “X”.
Construct a simple chart of weight, moment arms and torque moments as shown below, and place the values in each space.
Item A has a weight of 170 pounds. From the datum reference point, “A” has an arm length of 65 inches. Sixty five inches times 170 lb is 11050, so “A” has a torque moment of 11050 inches-pounds.
B weighs 340 pounds and its arm length is 174 inches. Since 340 times 174 equals 59160 inches-pounds, B has a torque moment of 59160 inches-pounds.
The total weight is 510 pounds and the total torque moment is 70210 inches-pounds, the balance point is found by dividing the total moment by the total weight.
This value is 137.6 inches located to the right of the Datum.
To check these values, let the moment arm of “A” be negative;
A's torque = force x distance (Left from CG)
- 170 pounds x -72.6 inches = −12353 in-lb
B's torque = force x distance (Right from CG)
- 340 pounds x 36.4 inches = 12353 in-lb
Since the counter clockwise torque (-12353) and the clockwise torque (12353) sum to zero, the helicopter will balance at the fulcrum point located 137.6 inches to the right of the Datum, point “X”.
All the calculations for the Bell206L II will be positive numbers due to the datum been out front of the helicopter.
Fuel Burn and CG
Fuel is the one weight that changes the CG in flight. It is critical to calculate CG correctly while the fuel is burnt off to mantain controlability of the aircraft.
The feul tanks of the B206L-1 Long Ranger II are the mid and aft seats.
Looking at the CG of full fuel tanks we find they are setup in an aft CG state. As fuel burns off we find the aircraft becomes lighter and the CG moves forward to its most forward position at 425lb. Then as it continues to burn-off it reaches its most aft position with 270lb of fuel in the tanks. From this CG point any fuel burn-off is forward until the fuel tanks are empty.
We can see the fuel CG movement in the table from the handbook on page 29.
And we can see the fuel CG movement when plotted on the graph envelope above from page 6 and as demonstrated in the example below.
These examples show what’s required to calculate the three hypothetical CG questions for the Longranger.
- Find the maximum Forward and Aft CG movements for a flight (due to fuel burn).
- Find the maximum or minimum weight that can be added or removed from the aircraft to remain within CG limits.
- How much weight must be moved to attain CG limits.
Forward and Aft CG Calculations
Find the maximum Forward and Aft CG movements with the load below:
- EW 2250lb
- Oil 13lb
- Pilot Row 1 Seat 1 170lb
- Passenger Row 2 Seat 3 185lb
- Passenger Row 3 Seat 5 192lb
- Passenger Row 3 Seat 7 136lb
- Baggage 42lb
- Take Off Fuel Full
- Fuel Burn to Empty Tanks
Above are three fuel tank pictures representing the different fuel levels relevant to the fuel table maximum CG shifts. In the example we are calculating the maximum forward and aft CG movements caused by the fuel burn-off.
After calculating the CG movements we can see that the maximum forward CG is 124.7 inches and the maximum AFT CG movement is 126.5 inches due to fuel burn. All other fuel burn CG movements are less extreme while remaining central.
Locker CG Calculations
We can see from our previous example during the entire flight the CG remained within allowable limits as shown on the graphed envelope. Now we are going to find how much weight can be added to the locker without exceeding the CG envelope limits.
Using the previous example we can see that full fuel and aft fuel loadings were the critical points in the flight, moving the CG closest to the right hand edge of the envelope (aft), all other fuel CG changes remained further inside the envelope.
Now lets calculate the CG using full fuel and the maximum amount of cargo that can be loaded in the locker, and then calculate again using aft fuel to demonstrate how weight in the locker will bring the CG to the edge of the graphed envelope.
By plotting them we can see both are outside the envelope. However if we draw a line from outside with locker weight to the inside without locker weight we notice the line cuts the edge of the envelope and that's the maximum amount of weight that can be added to the locker safely.
To prove this weight simply recalculate using the new weight in place of the maximum locker weight.
Of course you can use any amount of weight to be placed in the locker and add it to our Most AFT CG total in the table, then plot the new CG to find where it crosses the edge of the envelope. Read the difference between where the extended line crosses the edge of the envelope from the full fuel and aft fuel position.
But remember the structure limitation for the locker is 250 pounds maximum and in this example there is already 42 pounds in the locker leaving only 208 Lb to place in the locker.
We find full fuel is the limitation allowing only 85 Lb to be placed in the locker. If we took off with less than full fuel and burnt through aft fuel then aft fuel is the limitation allowing 100 Lb to be placed in the locker.
Using the previous example assume 250 lb was added to the locker. How much weight needs to be moved from the locker to the vacant front seat to bring the CG back into the edge of the envelope?
- Use aft fuel and load the maximum amount of weight in the locker to find how far outside the edge of the envelope the CG is.
- Now choose any weight from the locker and place it on the front seat. Recalculate the CG and plot it. I decided to use 30 pounds.
- Calculate how many inches horizontally from outside the envelope to the new plot and compare it to the pounds. 129.4 – 128.4 = 1 inch. Find the ratio by dividing the pounds by the inches. i.e. 30 ÷ 1 = 30. For every 30 lb the CG moves 1 inch.
- Now calculate how many inches horizontally it is to the edge of the envelope from outside the envelope and multiply it by the ratio to find how much weight to move. 129.47 – 127.81 = 1.6 inches x 30 = 50 pounds.
- To prove this is correct recalculate and plot using the new weight.
Acculy you can use any gross weight for the helicopter, it won't affect the ratio of 1 inch = 30lb for locker to front seat weight shift.