Episode 165 – Safety, Security and the Science of G’s

Ensuring the passengers’ safety and security requires the driver to have the knowledge, skill, and experience to control the vehicle when confronted with an emergency. The emergency does not necessarily need to be a security scenario; it can often be an accident-producing situation. 

As we have mentioned many times in the past, research and science define driving skill as the driver’s “ability” to use the vehicle’s “capability.”

Former VDI or Scotti School students know that we were and are anal about training our students to maximize the vehicle’s capability. The more of the vehicle’s capability the security driver can use, the higher the probability that the driver will avoid an accident or other life-threatening scenario. 

To pass the VDI course or be certified as an ISDA Security Driver, a driver must demonstrate they can use a minimum of 80% of the vehicle’s capability.

The simple scientific fact is that every tenth of a percent that VDI can train the driver’s ability to use the vehicle’s capability is a lifesaving skill. But it is hard to conceptualize how a 1% increase in driver ability increases the principal and passengers’ safety and security; hence, an explanation is in order.

Safety/Security and the Concept of G’S

We need to explain the G forces’ effect on the Driver/Principal/Vehicle safety and security.

The best way to understand a specific vehicle’s safety dynamics is by understanding how G-forces affect the driver/vehicle’s capability to stay in control. Anytime the steering wheel is moved while the car is in motion, a lateral or sideways force is created. This force is pushing in the opposite direction the car is turning. 

The term G’s is a measurement of the force acting on the car. It is this force that determines if the driver can stay in control of their vehicle. A tenth of a G can make the difference between avoiding a potential accident or vehicle attack scenario – or not.

Scenario A – at a given speed, if the path of a 4,000 Lb/1814.4 Kg car is altered in a way that produces one G, that one G of force is equivalent to 4,000 Lb/1814.4 Kg of force pushing on the vehicle center of gravity (CG) pushing the vehicle away from its desired path.

There is a simple equation that is used to determine the amount of force pushing on the vehicle. The amount of force pushing on the car is equal to the G’s times the vehicle’s weight.  

Scenario B – At that same given speed in scenario A, if we turned the steering wheel of our 4,000 Lb/1814.4 Kg car in such a way that .7G was created, then we would have created 2,800 lbs./1270 Kg of force. We came to those numbers by multiply the vehicle weight times .7 G’s or 4,000/1814.4 Kg X .7 g’s = 2,800 lbs./1270 Kg.  

If the car weighed 3,000 lbs/1361 Kg and the steering wheel was turned the same number of degrees at the same speed, and the vehicle took the same path, the equation would read 3,000 lbs/1361 Kg (the vehicle weight)  X .7 g’s or 2,100 lbs/953 Kg of force, and so forth. 

Understand the lateral G-forces created in a turn are based upon both the vehicle’s speed, weight, and how much the steering wheel is moved or the degree, or sharpness, of the turn.

The purpose of using G’s as a measuring tool

G’s are used as a measurement of driving skills and method of testing vehicle capability. The purpose of using G’s instead of weight for a measuring tool is because saying that “there are 2,800 lbs./1270 Kg. of force being exerted on the car” does not tell us very much.  

The 2,800 lbs./1270 Kg. of force number is meaningless unless the weight of the car is also known. For instance, if 3,000 lbs/1361 Kg of force is exerted on a 5,000 lb/2268Kg car, that’s no big problem. 

However, if you take a corner in such a way that 3,000 lbs/1361 Kg. is being exerted on a 2,000 lb/907 Kg car, you’re in big trouble. 

It’s far easier to say, “This vehicle is designed to absorb .7 Gs.” If we use a 5,000 lb/2268Kg car for this example, then .7 Gs means the vehicle can absorb 3,500 lbs/1588 Kg of force before becoming unstable. (5,000 lb/2268Kg x .7 g’s) 

If we use our lightweight 2,000 lb/907 Kg car, then that .7 Gs is the equivalent of 1,400 lbs./635 Kg of force. (2,000 lb/907 Kg x .7 g’s) 


Now to the original question

 Why for decades have we been adamant about ensuring students reach 80% or more of the vehicle’s capability, and while conducting training, the instructors will spend an inordinate amount of time ensuring you can use every tenth of a G the vehicle offers?

The answer – a tenth of a G could mean the difference between surviving or not surviving. Here is an example:

Using the science of driving and applying metrics and equations that are familiar and often used in accident reconstruction technology, we can conceptualize the value of adding .1 G to the driver skill level.

If the vehicle is traveling 40 MPH/64.4 KPH in a vehicle/driver combination that can handle at .85 G’s, and there was an object (barrier) that measures 10 Feet/3.1 Meters wide in front of the driver, they would need approximately 49 Feet/15 Meters of distance to clear the barrier.

With all the parameters the same as above, such as vehicle position and speed (40 MPH/64.4 KPH) with a Vehicle/Driver combination that handled at .7 G’s, The driver would need approximately 55 Feet/17 Meters of distance to clear a barrier that 10 Feet/3.1 Meters.

If traveling 40 MPH/64.4 KPH with a Vehicle/Driver combination that handled at .6 G’s, the driver would need approximately 59 Feet / 18 Meters distance to clear a barrier that is 10 Feet/3.1 meters.

Same scenario traveling 40 MPH/64.4 KPH but now with a Vehicle/Driver combination that handled at .5 G’s the driver would need approximately 65 Feet/19.8 Meters distance to clear a 10 Feet/3.1 Meter barrier.

With all the parameters the same as above, such as vehicle position and speed (40 MPH/64.4 KPH) but now with a Driver/Vehicle combination that can handle only .4 G’s, they would need approximately 72 Feet/ 22 Meters of distance to clear the same barrier.

Every 10th of a G that you can use gives you more time and distance to drive out of the problem – this is why computers are needed to measure the driver’s ability to use the vehicle’s capability.

These measurements can be computed using a radar gun and accurately measuring the vehicle’s path through individual exercises. 

Understanding the concept of lateral acceleration is a must for a security driver.

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