Conversion of g acceleration into "body weight"

[tjvv]

Hi guys,

I have a table with vertical peak acceleration values [g] and I want to find out a formula that convert the [g] values into ground reaction forces in units of body weight (BW). I think Newton´s 2nd law (F = m x a) may help but I am stuck on finding a generic relationship (problem is that I do not have the body mass values...)

Example: peak acceleration of a walking person is 2 "g". How much is it in terms of body weights?

Activity [g] [Body weight]
-----------------------------------------------------------
Walking 2 ?
Jumping 5 ?Thank you!

tjvv

 

[A.T.]

I have a table with vertical peak acceleration values [g]

Coordinate acceleration (CA) relative to ground, or proper acceleration (PA) that an accelerometer measures?

and I want to find out a formula that convert the [g] values into "body weights".

You mean the vertical ground reaction force (VGRF) needed to achieve that vertical acceleration?

VGRF[BW] = PA[g] = CA+1[g]

 

[Andrew Mason]

Hi guys,

I have a table with vertical peak acceleration values [g] and I want to find out a formula that convert the [g] values into ground reaction forces in units of body weight (BW). I think Newton´s 2nd law (F = m x a) may help but I am stuck on finding a generic relationship (problem is that I do not have the body mass values...)

Example: peak acceleration of a walking person is 2 "g". How much is it in terms of body weights?

Activity [g] [Body weight]
-----------------------------------------------------------
Walking 2 ?
Jumping 5 ?Thank you!

tjvv

Body weight is a force so to express body weight in terms of acceleration you would have to multiply by the body's mass.

g = 9.8 m/sec². The body weight of a person of mass M (in kg) is Mg =9.8M Newtons.

So a person walking horizontally and accelerating horizontally at 2g = 19.6 m/sec² would have to exert a force of 2Mg = 19.6M Newtons, where M=mass of the person.

AM

 

[tjvv]

Coordinate acceleration (CA) relative to ground, or proper acceleration (PA) that an accelerometer measures?

I mean proper acceleration (PA) that an accelerometer measures.

You mean the vertical ground reaction force (VGRF) needed to achieve that vertical acceleration?

Yes, when I mean bodyweight it is the vertical ground reaction force.

Since it is a force we can put in Newton´s 2nd law (F=m * a):

F = [(m * acceleration measured) / (m * Earth acceleration)] + 1 //+1 is to consider only linear acceleration (excluding Earth gravity)
BW = [acceleration measured / Earth gravity] +1

So in the example from the table above:

BW = [2 / 10] +1 //considering gravity as 10m/s2 to simplify
BW = 0.2 +1
BW = 1.2 which means the force of walking would be 1.2 times the body weight of a person

General formula would be this: BW = [acceleration measured / Earth gravity] +1

Is this correct?

 

[A.T.]

I mean proper acceleration (PA) that an accelerometer measures.
...
General formula would be this: BW = [acceleration measured / Earth gravity] +1

Is this correct?

If "acceleration measured" is proper acceleration then you don't need that "+1". See my formula in post #2.

 

[tjvv]

Hi AT,

sorry I meant Coordinate acceleration (CA) relative to ground (since it excludes the Earth gravity).

So from your formula it means that to convert a G force into BW it is just sum the measured g force with earth´s gravity?
Example:
BW = CA+1[g]
BW = 2 + 1
BW = 3 (getting the initial table example, would mean that the force acting when people is walking would be 3 times person´s body weight)

Can you please confirm?

Thanks

 

[A.T.]

Hi AT,

sorry I meant Coordinate acceleration (CA) relative to ground (since it excludes the Earth gravity).

So from your formula it means that to convert a G force into BW it is just sum the measured g force with earth´s gravity?
Example:
BW = CA+1[g]
BW = 2 + 1
BW = 3 (getting the initial table example, would mean that the force acting when people is walking would be 3 times person´s body weight)

Can you please confirm?

To accelerate your center of mass at 2g upwards, relative to the ground, you have to apply 3 times your body weight to the ground. But you don't have such high accelerations during walking.

 

[Andrew Mason]

Hi AT,

BW = 3 (getting the initial table example, would mean that the force acting when people is walking would be 3 times person´s body weight)

Just to add to what AT has said, if you think of the acceleration provided by a 100 m. sprinter running the 100 m. dash in 10.4 seconds and taking 1 second to get up to a speed of 10 m/sec, the acceleration is only 1 g. I think that is pretty close to the maximum horizontal human acceleration you are going to see.

AM

 

[A.T.]

Just to add to what AT has said, if you think of the acceleration provided by a 100 m. sprinter running the 100 m. dash in 10.4 seconds and taking 1 second to get up to a speed of 10 m/sec, the acceleration is only 1 g. I think that is pretty close to the maximum horizontal human acceleration you are going to see.

He is asking about vertical accelerations, which can be higher for a fraction of a second during landing impact (jumping, running fast). But during normal walking the vertical ground reaction doesn't go much beyond 1 body weight, so the vertical acceleration is a small fraction of 1g.