B Understanding G-Force: x,y and z Components Explained

  • B
  • Thread starter Thread starter solarmidnightrose
  • Start date Start date
  • Tags Tags
    G-force G-forces
AI Thread Summary
G-Force has x, y, and z components that represent different directions of acceleration experienced in a moving vehicle. The x-axis typically corresponds to side-to-side movements, the y-axis to forward and backward forces, and the z-axis to vertical forces, such as gravity. Understanding these components involves recognizing that G-Force acts as a vector, with both magnitude and direction, influenced by the reference frame of the observer. Calculating the total G-Force requires using the formula g = √(gx² + gy² + gz²) and may involve trigonometry to determine the direction. The discussion emphasizes the importance of the reference frame when interpreting G-Force data from accelerometer apps.
solarmidnightrose
Messages
28
Reaction score
10
TL;DR Summary
What is G-Force?
I've been to the amusement park recently, and one day I decided to use an app (Physics Toolbox), to collect some data.

I downloaded the data, and it gave it to me in a .csv format. This table of data showed that there were x,y and z values for G-Force.

I don't understand how G-Force has x,y and z components/aspects?
Please help me to understand this.

Thank you.
 
Physics news on Phys.org
Think about being in a car that's going up an incline, around a turn, and accelerating all at the same time.
 
phinds said:
Think about being in a car that's going up an incline, around a turn, and accelerating all at the same time.

Okay, so I've thought about what you've said... and this is how my brain interprets it:
The car is doing 3 different things all at once... If G-force is like the car that can do three things at once, then that would result in it having an x,y and z component.​
G-Force can travel in different directions? (is that what the message is?)​
Well, thank you phinds for your input, but I still don't understand what x/y/z components of G-Force actually are? In what direction do they act?

Also, to calculate the total G-Force, how would you do that? Would you have to use Trigonometry or is there some special formula specified for G-Forces?

Thanks.
 
solarmidnightrose said:
but I still don't understand what x/y/z components of G-Force actually are? In what direction do they act?
It all depends on what 'reference frame' you are using. Your App is traveling with you its reference frame will be a local one and, if you line it up so that its screen faces forwards and the bottom is towards the floor of the car then those are the xyz directions. That would be the easiest frame in which to consider your ride. When you are stationary, only the z (down to the floor, say) would have a g value (giving you your weight). As you accelerate in a straight line, the y (fore and aft) would start to show a backwards g value and when you enter a bend, you will have a g component towards the outside of the curve. The signs of these accelerations can be confusing and they need to be consistent amongst themselves and with how you 'feel' them. You 'feel' a centrifugal force because the car is pushing you into the curve etc..
The overall g force magnitude would be g = √(gx2 + gy2 +gz2) and the direction would be found by Trigonometry. That involves specifying the angles in your Cartesian xyz axes and you can think of it in terms of Spherical Polar Co ordinates - which you may or may not want to get into. This link will do it for you (you enter the xyz co ordinates into the box at the top and the spherical values turn up there. The diagram at the bottom of the page tells you how the angles are defined. Try some simple values and check that the results make sense to you.

PS Your data will not be consistent with the above unless you happened to be holding the phone the way I describe but, if y ou held the phone steady, you could use the above calculator to tell you which way is 'up' when you started off.
 
Last edited:
solarmidnightrose said:
Summary: What is G-Force?

I don't understand how G-Force has x,y and z components/aspects?
Please help me to understand this.

Do a google search for vector components.
 
  • Like
Likes davenn and sophiecentaur
solarmidnightrose said:
Well, thank you phinds for your input, but I still don't understand what x/y/z components of G-Force actually are? In what direction do they act?
This is an app on your phone, right? The accelerometers are fixed to the phone. On my Samsung, when holding the phone flat, x is to the sides, y is forward/back and z is perpendicular to the screen.
 
  • Like
Likes solarmidnightrose and sophiecentaur
@solarmidnightrose
Do you remember doing the parallelogram of forces in your youth? That's simple vectors in 2D. It extends to 3D in the same way.
 
solarmidnightrose said:
Summary: What is G-Force?

I've been to the amusement park recently, and one day I decided to use an app (Physics Toolbox), to collect some data.

I downloaded the data, and it gave it to me in a .csv format. This table of data showed that there were x,y and z values for G-Force.

I don't understand how G-Force has x,y and z components/aspects?
Please help me to understand this.

Thank you.

If you push an object, it accelerates in the direction you are pushing it. Now, suppose that you are sitting in a car with your back against the seat. The car gets a push from behind (say, another car hits it). Then the car goes forward. But you don't--at least, not immediately. Because nobody is pushing you. But the seat that you're sitting in is attached to the car, so it moves forward, pushing against your back. So you feel the seat pushing against you. Your brain, which is used to your being held down by gravity interprets what's happening as some kind of artificial gravity pulling you back into the seat. But that's not what's really happening---it's really the seat pushing forward into you.

From your point of view, you can think of the car as at "rest" and there is a weird kind of fake gravity pulling you back into the seat. That fake gravity is what people call "g-forces". It's just the opposite of the force of the seat pushing against you, just re-interpreted by your brain as fake gravity pulling you back into the seat.

It's a vector; it has a direction and a magnitude. The direction is the opposite the direction your car is being shoved.
 
  • Like
Likes sophiecentaur
sophiecentaur said:
It all depends on what 'reference frame' you are using. Your App is traveling with you its reference frame will be a local one and, if you line it up so that its screen faces forwards and the bottom is towards the floor of the car then those are the xyz directions. That would be the easiest frame in which to consider your ride. When you are stationary, only the z (down to the floor, say) would have a g value (giving you your weight). As you accelerate in a straight line, the y (fore and aft) would start to show a backwards g value and when you enter a bend, you will have a g component towards the outside of the curve. The signs of these accelerations can be confusing and they need to be consistent amongst themselves and with how you 'feel' them. You 'feel' a centrifugal force because the car is pushing you into the curve etc..
The overall g force magnitude would be g = √(gx2 + gy2 +gz2) and the direction would be found by Trigonometry. That involves specifying the angles in your Cartesian xyz axes and you can think of it in terms of Spherical Polar Co ordinates - which you may or may not want to get into. This link will do it for you (you enter the xyz co ordinates into the box at the top and the spherical values turn up there. The diagram at the bottom of the page tells you how the angles are defined. Try some simple values and check that the results make sense to you.

PS Your data will not be consistent with the above unless you happened to be holding the phone the way I describe but, if y ou held the phone steady, you could use the above calculator to tell you which way is 'up' when you started off.
Yes, this is all beginning to make sense to me now. Once I clicked the link you attached it all began to fall into place.
Thank you
 
  • #10
russ_watters said:
This is an app on your phone, right? The accelerometers are fixed to the phone. On my Samsung, when holding the phone flat, x is to the sides, y is forward/back and z is perpendicular to the screen.
Yep this is an app on a phone. And I get the same results as you have stated.

Thanks.
 
  • Like
Likes russ_watters
  • #11
stevendaryl said:
If you push an object, it accelerates in the direction you are pushing it. Now, suppose that you are sitting in a car with your back against the seat. The car gets a push from behind (say, another car hits it). Then the car goes forward. But you don't--at least, not immediately. Because nobody is pushing you. But the seat that you're sitting in is attached to the car, so it moves forward, pushing against your back. So you feel the seat pushing against you. Your brain, which is used to your being held down by gravity interprets what's happening as some kind of artificial gravity pulling you back into the seat. But that's not what's really happening---it's really the seat pushing forward into you.

From your point of view, you can think of the car as at "rest" and there is a weird kind of fake gravity pulling you back into the seat. That fake gravity is what people call "g-forces". It's just the opposite of the force of the seat pushing against you, just re-interpreted by your brain as fake gravity pulling you back into the seat.

It's a vector; it has a direction and a magnitude. The direction is the opposite the direction your car is being shoved.
A wonderful explanation I never would have thought of all by myself. I now understand this new concept of g-forces.

Thank you @stevendaryl
 
  • Like
Likes russ_watters
  • #12
sophiecentaur said:
@solarmidnightrose
Do you remember doing the parallelogram of forces in your youth? That's simple vectors in 2D. It extends to 3D in the same way.
Hi @sophiecentaur I have only begun learning about 2D/3D vectors earlier this year, so this is all very new to me. Nevermind that-this is very good revision for me.

Thanks :)
 
  • #13
solarmidnightrose said:
Hi @sophiecentaur I have only begun learning about 2D/3D vectors earlier this year, so this is all very new to me. Nevermind that-this is very good revision for me.

Thanks :)
The notion of Components of a Vector can be a hard one if it's not presented in the right way; resolving forces into components is somehow harder than combining two forces. So - one thing at a time. :smile:
 
  • Like
Likes solarmidnightrose

Similar threads

Replies
15
Views
3K
Replies
15
Views
3K
Replies
8
Views
1K
Replies
8
Views
2K
Replies
5
Views
2K
Replies
10
Views
2K
Back
Top