with regard to deployment. Determining compatibility is more than simply determining the volume compatibility of a canopy to a container size. Other factors, that need to be considered are the deployment type, technical standard order (TSO) certification, and placard limitations. 

## Reserve Bag Extraction Force 

When we ask the question, “How do you determine compatibility between approved parachute components,” the answer is, “Rig functionality must not be compromised.” Some say if a reserve canopy is too bulky, the reserve deployment bag is not easily extracted. If you can get the bagged canopy in the container, it should take no more than 18 pounds of pull force to extract it. 

## TSO Certification and Placard Limitations 

This area is one where many riggers have some confusion. According to AC 105-2, Sport Parachute Jumping, “the strength of the harness must always be equal to or greater than the maximum force generated by the canopy during certification tests.” In the case where the harness is certificated under TSO-C23b and the canopy under TSO- C23c, the maximum generated force of the canopy must not exceed the certificated category force of the harness and container (i.e., Low Speed Category (3,000 pounds) and Standard Category (5,000 pounds)). In this instance, no additional marking on the container is necessary. 

In the case where the canopy is certificated under TSO-C23b and the harness under TSO-C23c, the strength of the harness must be equal to or greater than the certificated category force of the canopy. 

For the current TSO-C23d, the peak force measured during the strength drops must be placarded on the outside of the harness. In this case, the strength of the canopy must not exceed that of the harness. 

The rigger, when making the determination as to whether a particular canopy and rig combination is compatible, must consider all of the above areas. If there is any doubt, the rigger should contact the rig manufacturer for guidance. 

## Harness Strength 

TSO-C23b was originally written back in the 1940s before the advent of square parachutes. It had two categories under which a parachute system could be certified. The “Low Speed” category was limited to use in aircraft under 150 miles per hour (MPH) and certified to 3,000 pounds. This category required large block letters decrying “limited to use in aircraft under 150 MPH.” It also had a “Standard Category.” This category required no warning labels and had neither weight nor speed limitations and was tested and certified to 5,000 

pounds. It is important to note that neither category had a weight limitation. 

Weight has only a minimum effect on parachute opening forces. To be exact, if you were to increase a given weight by 50 percent, you would only see a 5 percent increase in opening force; likewise if you double that given weight, you would only see a 10 percent increase in opening force. 

This seems counter intuitive until you think about it. Speed is the critical factor that hurts us and our equipment when we have the occasional hard opening. But because speed is often derived from mass or weight, we associate the hard opening with primarily weight. Let us look at the calculations. 

The math model for opening forces is described in the “Recovery Systems Design Guide” by Theodore Knacke. The definitions and formula is as follows: 

• Force—total opening forces 

• C$_{d}$ —drag coefficient of canopy 

• S$_{o}$ —square footage of canopy 

• Q—dynamic pressure in pounds per/square foot (½ ρʋ $^{2}$) 

• X$_{1}$—decreasing load factor 

NOTE: There are 2 methods for deriving this factor. The Pflanze method and a lookup of the chart included in the reference manual. The chart is the simplest for personnel parachutes and effectively results in being one tenth of the pound per square foot loading. 

• C$_{x}$ = shock load coefficient which is derived from testing and includes such things as slider size, brake setting, angle of nose cut, etc. For this exercise we will use a value of 1 as this number ranges from .5 to 1.5 or so. Without a slider it can go as high as 10. 

Therefore: Force = C$_{d }$ × S$_{o}$  × Q × X$_{1}$  × C$_{x}$ 

If we group the C$_{d}$  × S$_{o}$  × Q and calculate them, at first we get a big number ie: 

C$_{d}$ = .8 S$_{o}$ = 200 square feet Q = 33 PSF @ 117MPH together = .8 × 200 × 33 = 5,280 pounds. This number is then ameliorated by the X$_{1 }$ decreasing load factor and the C$_{x}$ shock load factor. If the C$_{x}$ is 1 (and we will assume this for this example), then it has no effect on the outcome. 

The X$_{1}$  factor is the key because it is based on pounds per square foot loading multiplied by 1. If you try different weight values and reiterate the formula, you can see it only 

