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Is someone know how to calculate vertical G-force

  1. Oct 22, 2011 #1
    I eager want to know about how to calculate the vertical G-force.I have some ideas but I do not if it is right or not.I just calculate the acceleration first,then use a/9.8 kg/N or m/s^2.

    Is this formula right or not?Is someone can tell me?

    Thanks in advance.
  2. jcsd
  3. Oct 22, 2011 #2
    you want to calculate G, the universal gravitational constant.....

    (1) F = GmM/r2

    (2) So hang some mass on a scale.....F (weight) = ma =mg......

    and plug m in (1) where you know all the pieces except G. (M is the mass of the
    earth at about 5.9722 × 1024 kg and r it's radius is 6371 kilometers.
  4. Oct 22, 2011 #3
    You mean mg=GmM/r^2,g=GM/r^2.Usually,g=9.8kg/N=9.8m/s^2.Right?
    The situation is that when a 4.5Tons machine put onto the shock absorbers(x4).How to calculate the machine's vertical G-force during shipping.I calculated the acceleration of the vertical a and use a/9.8.It means "a/9.8" is the G-force of the machine.Is this concept right or not?
  5. Oct 22, 2011 #4
    m/s[itex]^{2}[/itex] is not same as kg/N
  6. Oct 22, 2011 #5


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    Not sure what you are trying to measure. What is 'a' again??? a is the acceleration of the vertical???
    If a machine that weights 4.5 Tons is based on 4 shock absorbers then each shock absorber gets about 4.5/4=1.16 Tons of weight if the machine doesnt move at all.
  7. Oct 22, 2011 #6
    Thanks for everyone's help.
    Sorry,m/s^2 the same as N/kg caused when you use different unit.
    "a" is the acceleration(vertical and horizontal).Yes,the 4.5 Tons is based on 4 shock absorber then each shock absorber gets about 4.5/4=1.125 Tons of weight.
    When the truck went through a sloping or a hole,the shock absorber will be pressed and I just consider that the displacement is the max data(-5mm) during shipping.At this time,there is a vertical acceleration we call it "a".There is a explain I found that "Is a surface of the Earth gravity, 1G in aviation is defined as the aviation flight at sea level and by the gravity lift and down to attract the power of phase equilibrium".That's why I just use "a"(acceleration)/9.8 to calculate the G-force.But I am not sure if it is right or not.
  8. Oct 22, 2011 #7
    Hi Gyao. I am reading all the threads in your post and this is what I understand:
    1. You have a machine that weights 4.5 Tons.
    2. The machine is sitting on 4 shock absorbers.
    3. You are trying to calculate how many G's your machine will be subject to during shipping.
    Is this correct?
    If this is the case, what you are trying to calculate is not as straight forward as you might think. But I think I can give you a couple of tips:
    Here is an example. Say you drop a piano (not your machine cause I imagine you machine is very expensive) from a 2 story building. During the fall the piano is subject to 1G ( but the piano is happy during the fall and feels zero :smile: ). When it hits the ground the piano will decelerates in a very short time and the piano will be subject to a very large number of G;s. The total number of G’s depends of the height of the building and the deceleration time during the impact.
    Similarly if your machine goes through a very large bump, the G;s that the machine will feel depend on the height of the bump, the maximum suspension travel of your shocks, and the properties of your shock absorbers.
    You have 2 options:
    You can make an educated guess , but you will need the right information and make some assumptions . For example what is the maximum height drop you think your machine will encounter during the shipping? Or, when you bought your shock absorbers, did the manufacturer provide you with a spring or damping rate N/m or N/m.s or a chart? Or a maximum suspension travel?
    Option 2 will be to measure the G;s experimentally with an accelerometer. Expensive equipment but the most accurate solution.
    Also you might want to look into shipping labels that bleed when the crate has been subject to certain G’s. Buy different levels and place them inside and outside the crate. Ahh and insure your shipment!
    Good luck
    Last edited: Oct 22, 2011
  9. Oct 23, 2011 #8
    Shock absorbers don't support any weight, they resist motion, with a force opposing the motion, from memory, of something like F = -kv*v. (sorry too lazy to figure out how to type v squared).

    Springs support weight, and again from memory, something like F= -S(x1-x0).

    The springs in my car are for example 400 lb/in, and my 3200 lb car compresses each of the springs on the 4 wheels about 2 inches when not moving.

    What are you trying to do?
  10. Oct 23, 2011 #9


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    Ok i see now, provided that he knows the acceleration 'a' at which the 4.5 tons is moving down (though i didnt understand how he measures it) he wants to know how many Gs this acceleration is equivalent to. Acceleration of 9.8m/s^2 is equivalent of 1G, so yes 'a/9.8' is how you ll find out how many Gs your acceleration is but you have to be sure that you measured the acceleration 'a' in units of m/s^2. 1m/s^2=1Newton/1Kg
  11. Oct 23, 2011 #10
    So,that's the G-forces calculation,right?I just need to make sure the units of acceleration.The "a" I mentioned to just calculate as a=(kx-mg)/m."k" and "x" provided by supply."m" is the weight of the machine and "g=9.8N/kg".k=F/x=1700x9.8/-5(mm)=-3332.So,a=((-3332)*(2.365-5)-1125*9.8)/1125=2.So the G-force=2/9.8=0.2G. Is this right?
    Last edited: Oct 23, 2011
  12. Oct 23, 2011 #11
    Yes,you are right.I know how large the Maximum weight of the shock absorber can support about 1700kg and I know the shock absorber moving down as limit about -5mm.
  13. Oct 23, 2011 #12
    Lack of information on what is actually being attempted makes me predict failure. 4.5 tons will need professional attention to survive shipping.

    The term "shock absorber" continues to be used when whatever the actual device is, it won't be just a shock absorber, something else has to support the weight like a spring. The mass, spring rate, and damping factor need to match up to the expected handling to be effective.

    What will happen when your shipment is tilted say 30 degrees, or dropped 4 inches? Both events are typical in even the most fragile shipping.
  14. Oct 23, 2011 #13
    Yes,you are correct.I considered the tilted and fastend the machine onto the truck.The machine can not horizontal moving and just have a short vertical moving.My question is that I do not know the vertical G-force during shipping.

  15. Oct 23, 2011 #14
    Thanks for everybody's help.
    I learned more from the forum.I want to know vertical G-force calculation."acceleration/9.8"is the G-force or it's wrong.
    Thanks again.
  16. Oct 23, 2011 #15
    Sitting on the truck without the truck moving is 1 G. Once the truck is moving its gets much more complicated.
  17. Oct 24, 2011 #16


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    5mm have to be converted to meters(m), that is 5mm=0.005m. Are you sure that it is mm or is it cm? Cause it seems weird to me that the spring moves down only 5mm. Also what is 2.365?
  18. Oct 24, 2011 #17
    Yes,I am sure it is "mm".My purpose is to calculate the Max and Min G-force of the machine during shipping.The shock absorber is made from rubber and metal.When the machine put onto the 4 shock absorbers it will move down 2.365mm.I did not convert to the meters just because it can be ignore when I use the formula a=(kx-mg)/g.That's all my idea.The environment is complicated.I considered that when the truck went through a hole or a slope.

    What do you think?Do you think that's the biggest G-force or you have different ideas?Could you please help think about it and we can discuss it.I know it can not be very precision but it can tell us a probable data.
    Last edited: Oct 24, 2011
  19. Oct 24, 2011 #18


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    I think there is a mistake in the calculation. The term 2.365-5 should not be there. If the total compression of the spring is 5mm then the extra compression due to moving above a hole or a slope might be 2.365-5 *but* the total compression will be 5mm so the total force from the machine to the spring is k*5mm. So just put 5mm instead of 2.365-5 and do the calculation again it will give a=5,008 so the equivalent G is about 0.5G. The way you did it is like subtracting the weight mg two times.(and btw k cant be negative)
  20. Oct 24, 2011 #19
    Well.Thank-you Delta².Yes the "k" has to be plus.You know customer's request is the G-force of the machine from 1G to 6G.As one of forum friends said that it is 1G during shipping if the road is smooth without rush ,hole and slop.So when the truck went through a hole or a slop it seems that the G-force of machine should be larger than 1G.What do you think?I do not know why the G-force of the machine just 0.5G.It is unbelievable.

  21. Oct 24, 2011 #20


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    The question makes no sense. The maximum G it can experience is the same maximum G that the outer housing may experience. This is because the spring may bottom out and provide no suspension at some degree of acceleration.

    If you are asking 'what is the acceleration before the spring bottoms out' then you need to compress the spring to its full extent and measure how much force that requires.

    Now, you might be asking some more subtle question without recognising it, like what is the maximum shock loading the spring and damper system could impose on your kit, (whilst not bottoming out). If so, then we need to discuss the frequency spectrum of the shock load that you are trying to predict, and more details of the spring/damper system enough to describe its SHM behaviour (e.g. natural frequency and damping behaviour).
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