Thread: Force Calculations involving the velocity

Hi guys, I'm working on a little project and have come across a little problems in terms of calculating the force. I am implementing a device which takes recordings at set intervals of times and records movements (Accelereometer) in all 3 axis. This device is always constantly moving with various speeds, as were I get a little confused. Okay So this is the data I have and am recording....

Time (the time intervals are 10 milliseconds apart)
Displacement (in regards to accelerometer) (from intitial position)
Velocity (Speed in km/hr)

Can anyone help me with this? I've run a few things but I keep getting unknown variables in my equations. I know i'm missing somethign simple but I don't kno why.

Originally Posted by Samuel

Hi guys, I'm working on a little project and have come across a little problems in terms of calculating the force. I am implementing a device which takes recordings at set intervals of times and records movements (Accelereometer) in all 3 axis. This device is always constantly moving with various speeds, as were I get a little confused. Okay So this is the data I have and am recording....

Time (the time intervals are 10 milliseconds apart)
Displacement (in regards to accelerometer) (from intitial position)
Velocity (Speed in km/hr)

Can anyone help me with this? I've run a few things but I keep getting unknown variables in my equations. I know i'm missing somethign simple but I don't kno why.

You really have not asked a specific question so I don't know how anyone can answer you.

One thing though, if you really are trying to calculate a force, you have not indicated any mass or momentum readings, so I do not see how you will be able to do that.

If all you care about is calculating acceleration, well, isn't what the accelerometer is giving you in discrete time intervals?

hello,

Sorry and yes I am tryign to calculate the force that is being applied. Using the accelerometer I can calculate the Force by simply usings the readings. I do this by;

Equation;
The 0 and 1g values are gathered from the calibration process where the device is placed on all (+'ve & -'ve) axis's and the data is recorded.

Force = (1 / (difference of 0 and 1g)) x (Displacement of accelerometer).

The problem with this is I do not take into account acceleration / velocity what so ever, so If i graph my results I can not tell the difference between a bump at 40km/hr and 80km's/hr.

Originally Posted by Samuel

hello,

Sorry and yes I am tryign to calculate the force that is being applied. Using the accelerometer I can calculate the Force by simply usings the readings. I do this by;

Equation;
The 0 and 1g values are gathered from the calibration process where the device is placed on all (+'ve & -'ve) axis's and the data is recorded.

Force = (1 / (difference of 0 and 1g)) x (Displacement of accelerometer).

The problem with this is I do not take into account acceleration / velocity what so ever, so If i graph my results I can not tell the difference between a bump at 40km/hr and 80km's/hr.

There is a physics sub-forum here, and you might want to post this question there. I don't what problem you are trying to solve, but if you are moving with constant velocity, whether at 40 or 80 km/hr, your acceleration is 0. And if you then accelerate the body, i.e., apply a force in order to change its velocity, then unless you are working at relativistic speeds, the velocity of the object has no bearing on the change in velocity due to the acceleration. Whether the object is moving at 0, or 1, or 10 million km/hr, accelerating it by 1 km/hr does just that, changes its velocity by 1 km/hr.

I understand what you're saying but if I graph this using Force against Time a raise will be different depending on speed because of the time that has passed for the displacement to revert back to zero. Because of the application it is being implemented into it will not simply be a case of "Bump" then 0 again. It will be "bump" then a little period where it goes negative, positive, negative, positive etc to retrieve back to initial position. It obviously will not be a large "return to 0" period but it will be noticable in the graph.

would it not?