# Calculating approximate g-forces exerted on a surface. ### Help Support The Rocketry Forum: #### Serpico

##### Well-Known Member
I'm trying to calculate the ** approximate ** g-forces exerted on a surface. I know there are variables that will sway these calculations that may not be represented here (effect of the weight of the surface, drag, etc.).

Looking to get an approximation BUT willing to hear all arguments and critiques. This is my first shot at this:

Known factors:

Mass of object sitting on surface: 2.2410 kg
Max vertical acceleration of surface: 128.446 m/s/s
1 G-Force equals 9.8 m/s/s

My Calculations:

First I calculated the Force in Newtons exerted on the surface:
F = 2.241 * 128.4475968
F = 287.85 N

Now to calculate a single g-force exerted on the surface AT REST with the object sitting on the surface:
9.8 * 2.241
Force of 1 G on surface with this object = 21.9618 kg

Now convert that to g-forces exerted on the surface at max vertical acceleration:
G = 287.85 / 21.9618
G = 13.1068

So - if I did this correctly the surface experienced 13.1068 Gs based on the object weight of 2.241 kg moving at a max vertical acceleration of 128.446 m/s/s?

Last edited:

#### Rex R

##### LV2
er...I got the same result by dividing 128.446/9.8...
Rex

#### Serpico

##### Well-Known Member
er...I got the same result by dividing 128.446/9.8...
Rex

Sooo - Rex what you're saying is I got to the correct answer ~ just took the 481 mile route as opposed to the 1 step route?

#### Nytrunner

##### Pop lugs, not drugs

G-force is just the perceived "weight" of an object experiencing an acceleration that isn't gravity.
1 g is the standard acceleration imparted by earth's gravitational attraction (9.81 m/s^2 or 32.2 ft/s^2, pick your flavor and continue accordingly)
Since 1 g = 9.81 m/s^2 just like 1 dozen = 12 [items], its more of a descriptive than an actual unit.

Your 128 m/s^2 acceleration is ~13 g's. I don't know your end goal, but if its to calculate the resultant force on your surface, then I would skip the g's and just multiply your mass by your vertical acceleration like in Newton's 2nd law. (~288 N or ~65 lbs.)

My question is, why are you shoving a 5 lbs object upwards at ~420 ft/s^2?

#### Serpico

##### Well-Known Member
... My question is, why are you shoving a 5 lbs object upwards at ~420 ft/s^2?
Short story - I built a rocket, that carried a lander, that carried a rover. The lander was 5 lbs. It flew for about 3 seconds then CATO'd due to motor casing failure. My original calculations and notes are long lost. I'm putting together some facts about the project and needed a refresher RE the calculations. Thanks.

#### Nytrunner

##### Pop lugs, not drugs
Aw, that's terrible. Even when you have the physics right, quality control and reliability may still come back to bite you.

#### Rex R

##### LV2
kind of. before you can solve a problem you have to know what the problem is, or in this case what the question is. it seems to be human nature to over complicate things... , I just read through your post to find out what you're asking.
Rex

#### Rex R

##### LV2
actually now that I think of it we are both 'wrong' you gave the actual acceleration but were looking for total force acting on your payload...which includes the force of Earth's gravity so the answer would be a total force of 14.107 g
Rex