Hi Andrew
You ask is there a way to convert feet per second into G
Feet per second is a speed
A speed is always how long it takes to cover a distance.
G is not a speed it is an acceleration (or deceleration)
An acceleration (or deceleration) is how long it takes you to change speed.
To go from a speed to an acceleration (or deceleration) you need some more information
Stick with me and you should be able to work out any examples of this type that you need for your simulator.
You must be consistent with your units for speed. Feet per second would be best for your landing problem
Sorry if you know this next stuff!
Speed first:
Examples are metres per hour, metres per second, feet per minute, yards per year, inches per 10 years – all these are speeds because they have a distance covered divided by a time taken.
To change from one set of speed units to another is easy. Take the extreme one of inches per 10 years as an example:- divide the inches by 12 and you have changed the speed to feet per 10 years. Multiple the 10 years number by 10 and the speed becomes feet per year. Multiply that by 365 and the answer becomes feet per day. I’m sure you get the drift.
So lets say the speed was 60 inches per 10 years
that would become 60/12 = 5 feet in 10 years.
To get from feet per 10 years to feet per second you would have to divide that answer by 10 then 365 then 24 and finally 60. Clearly not many feet per second!
Now accelerations.
These are speeds divided by time. G is a rate of change of speed of 32.2 feet per second EVERY second. So if you jump off a cliff (ignoring air drag) thanks to gravity being termed 1G you will fall such that you pick up a speed of 32 feet per second for every second you are falling.
Now to your 747. As has been explained by Echo Tango you can convert your speed down the 3 deg ILS glide path (using trig) to a corresponding rate of descent (by the way useful speed unit conversions are that 45 kt = 76 feet per second while 60 mph = 88 ft per second)
So the big question is how do you convert this vertical speed of X feet per second into an acceleration or deceleration. The answer is YOU have to decide in what distance you are going to change your vertical SPEED from what it is on the approach to what it is after landing ie ZERO
This of course means making an assumption about how much the undercarriage legs will shorten due to the impact. That will vary from aircraft to aircraft, but I am guessing with a jumbo that from first touchdown ( of the rear tyres) to a fully compressed leg case with all tyres on the ground the centre of gravity of the aeroplane could have come down 4 feet (could be 6 feet I really don’t know)
Before we go the final lap, let me say that you must use your common sense with all these calculations about speed/distance/acceleration/time and so on. By that I mean understand whether the answer is going to be smaller or bigger when you make a change. This will tell you whether you have to divide by a factor or multiply by the same number.
Now the last lap. The greatest G when landing from a given rate of descent will be if the undercarriage of your aircraft only shortens a little bit. That means your vertical speed has to change VERY QUICKLY. This will feel like a very heavy landing and will mean lots of G
Let us say you decide your aeroplane has an undercarriage that shrinks 4 feet during the landing.
You have said you touchdown at 140 ft/min that is 140/60 = 2.3 ft/ sec
With these two “facts” we put these numbers in the appropriate equation of motion, which for feet per second units is
G = Vsquared/2H
Where G is the G of your touchdown (which we are tying to calculate)
V is YOUR declared touchdown vertical speed (must be in ft/sec)
H is the distance travelled in feet (the amount YOU decide the undercarriage squats)
So: the EXTRA G of touchdown for your numbers would be = 2.3 x 2.3 / 2 x 4 = 0.66
(a tad heavy according to bastud above)
Or if the undercarriage was a 6 ft job it would become 2.3 x 2.3 / 2 x 6 = 0.44
Any help?
JF