Rearranging pythagorus ...

[AbigailRocketblast]

I have 2 accelerometers mounted at right angles to each other.
I get a stream of data from each one going into my computer program. Ax and Ay.
Each accelerometer has a calibration factor to enable me to convert to engineering units. Fx and Fy.

Using Pythagorus I calculate the vector acceleration V.

V = sqrt ( (Ax.Fx)(Ax.Fx) + (Ay.Fy)(Ay.Fy) )

What I need to calculate is the equivalent calibration factor of the vector - ie Fv.

Initially I thought that it would simply be using Pythagorus and Fx and Fy, but this doesn't work. I thin started rearranging fomulas to try to work it out ...

we know that:

(Av.Fv)(Av.Fv) = (Ax.Fx)(Ax.Fx) + (Ay.Fy)(Ay.Fy)

same as:

(Av.Av)(Fv.Fv) = (Ax.Fx)(Ax.Fx) + Ay.Fy)(Ay.Fy)

and:

Av.Av = Ax.Ax + Ay.Ay

by combining those two formulas, is it possible to rearrange to solve for Fv?

This is what I have tried so far ... replacing Av.Av on the left

(Ax.Ax + Ay.Ay)(Fv.Fv) = (Ax.Fx)(Ax.Fx) + (Ay.Fy)(Ay.Fy)

then I kinda get lost ... am I on the right track?

Is this even possible?

Thank you.

 

[DrPhil]

I have 2 accelerometers mounted at right angles to each other.
I get a stream of data from each one going into my computer program. Ax and Ay. (m/s^2 or ft/s^2)
Each accelerometer has a calibration factor to enable me to convert to engineering units. Fx and Fy.
......Since you are using "F", I suppose that will be Force (N or lb). The conversion factor is a mass:
......F = m A

Using Pythagorus I calculate the vector acceleration V. Pythagorus will tell you the magnitude of a vector

V = sqrt ( (Ax.Fx)(Ax.Fx) + Ay.Fy)(Ay.Fy) )....|A| = sqrt(Ax^2 + Ay^2),....|F| = sqrt(Fx^2 + Fy^2)

What I need to calculate is the equivalent calibration factor of the vector - ie Fv.......|F| = m |A|

Initially I thought that it would simply be using Pythagorus and Fx and Fy, but this doesn't work. I thin started rearranging fomulas to try to work it out ...

we know that:

(Av.Fv)(Av.Fv) = (Ax.Fx)(Ax.Fx) + Ay.Fy)(Ay.Fy)

same as:

(Av.Av)(Fv.Fv) = (Ax.Fx)(Ax.Fx) + Ay.Fy)(Ay.Fy)

and:

Av.Av = Ax.Ax + Ay.Ay

by combining those two formulas, is it possible to rearrange to solve for Fv?

This is what I have tried so far ... replacing Av.Av on the left

(Ax.Ax + Ay.Ay)(Fv.Fv) = (Ax.Fx)(Ax.Fx) + Ay.Fy)(Ay.Fy)

then I kinda get lost ... am I on the right track?

Is this even possible?

Thank you.

I probably don't know what your variables really are. I expect V to be a velocity vector (Vx,Vy) with magnitude |V| = sqrt(Vx^2 + Vy^2)

Could you clarify? Please give definitions and units of A, F, and V.

 

[AbigailRocketblast]

I probably don't know what your variables really are. I expect V to be a velocity vector (Vx,Vy) with magnitude |V| = sqrt(Vx^2 + Vy^2)

Could you clarify? Please give definitions and units of A, F, and V.

Thank you DrPhil.

The accellerometers produce a value in distance/time^2 - that is they do not produce a figure in engineering units. However, each accelerometer has a calibration factor (F) which when applied to the value coming from the accelerometer gives us an accelerometer reading in m/s^2.

The Fx and Fy that I used was not related to force, but were the calibration factors for the accelerometers Ax and Ay respectively.

V is the magnitude of the vector obtained by applying Pythagorus to the outputs of Ax and Ay. If you can imagine 2 accelerometers at right angles to each other on a fixed object. If the object accelerates in both x and y at the same time, the magnitude of the combined acceleration will be V.

Does that help with rearranging the formula?

Thanks

 

[AbigailRocketblast]

Can you not do this:
a = Av, b = Fv, c = Ax, d = Fx, e = Ay, f = Fy

Then equation becomes:
a^2 b^2 = c^2 d^2 + e^2 f^2

Much easier to "handle"...

Thanks, Denis.

I can do that - actually I have - but it didn't help me to combine and rearrange the formula.

Have you had a go combining and rearranging? How did you get on?

Thanks