1. ## weight on waves

I have developed a scale to able use onboard a ferry. The ferry moves up and down, and usually 1 kg will show 0,7 – 1,3 kg on my application depending on the size of the waves. My solution is to build in 2 balance units. Balance 1 is not attached to anything but a 1 kg weight. Balance 2 is attached to the platform on the scale where to put the item. By knowing the actual weight on balance 1, I can read the output in my application. If fx balance shows 0,8 kg my conclusion is that the measured item on balance 2 is 20% to low. This works with small waves because I use a linear function. By knowing the actual weight and variation on balance 1, I need a formula including acceleration up or down and the gravitations influence in order to find the actual weight on balance 2. Is anybody able to figure this out?

2. Originally Posted by kaae
Balance 1 is not attached to anything but a 1 kg weight. Balance 2 is attached to the platform on the scale where to put the item.
Huh?

I can't figure out your setup so let's do this. I will show you how to calculate the apparent weight (aka the normal force, aka the scale reading) of an object on a scale under the influence of a sinusoidal wave.

According to Newton's 2nd the normal force on an object under an acceleration is
$\displaystyle N = ma + mg$
where m is the mass of the object.

Now, the position of a sinusoidally moving object can be represented as
$\displaystyle x(t) = A~sin( \omega t )$
where A is the amplitude of the wave and $\displaystyle \omega$ is the angular frequency of the wave. (In order to use this formula we must have that $\displaystyle x(0) = 0$ and $\displaystyle v(0) = \omega A$. You can see this from the formulas below. If you can't arrange these conditions then you will need to include a phase angle into the argument of the sine function.)

So
$\displaystyle v(t) = \frac{dx}{dt} = \omega A~cos( \omega t )$

$\displaystyle a(t) = \frac{dv}{dt} = -\omega ^2 A~sin( \omega t )$

Thus your normal force (scale reading) as a function of time is:
$\displaystyle N(t) = m( g - \omega ^2 A~sin( \omega t ))$

-Dan

3. ## Weight on waves

Inside the scale a 1 kg load is attached. I messure the variation on this and multiply the variation direct on balance 2. Balance 2 is the one attached the the platform where you put fx in my case a bottle of whiskey. If it shows 2 kg at the time where balance 1 shows 0,8 kg, I know that the 2 kg is 20% to little. Therefore I conclude that the item on the scale (balance 2) is 2,5 Kg. What do I miss in the formula besides the direct percantage variation?

4. Originally Posted by topsquark
Huh?

I can't figure out your setup so let's do this. I will show you how to calculate the apparent weight (aka the normal force, aka the scale reading) of an object on a scale under the influence of a sinusoidal wave.

According to Newton's 2nd the normal force on an object under an acceleration is
$\displaystyle N = ma + mg$
where m is the mass of the object.

Now, the position of a sinusoidally moving object can be represented as
$\displaystyle x(t) = A~sin( \omega t )$
where A is the amplitude of the wave and $\displaystyle \omega$ is the angular frequency of the wave. (In order to use this formula we must have that $\displaystyle x(0) = 0$ and $\displaystyle v(0) = \omega A$. You can see this from the formulas below. If you can't arrange these conditions then you will need to include a phase angle into the argument of the sine function.)

So
$\displaystyle v(t) = \frac{dx}{dt} = \omega A~cos( \omega t )$

$\displaystyle a(t) = \frac{dv}{dt} = -\omega ^2 A~sin( \omega t )$

Thus your normal force (scale reading) as a function of time is:
$\displaystyle N(t) = m( g - \omega ^2 A~sin( \omega t ))$

-Dan
and the time avaerge of $\displaystyle N(t)$ is $\displaystyle mg$ the weight.

RonL

5. Originally Posted by kaae
Inside the scale a 1 kg load is attached. I messure the variation on this and multiply the variation direct on balance 2. Balance 2 is the one attached the the platform where you put fx in my case a bottle of whiskey. If it shows 2 kg at the time where balance 1 shows 0,8 kg, I know that the 2 kg is 20% to little. Therefore I conclude that the item on the scale (balance 2) is 2,5 Kg. What do I miss in the formula besides the direct percantage variation?
Lets assume these are something like spring balances. How do you ensure that:

1. If identical they are (and remain) in phase?

2. How well do you think you will be able to jointly calibrate the scales?

RonL

6. ## Balances in phase?

The two balances is once calibrated with a 100% correct 2000g load. Both balances is attached to the same PLC and deliver results 60 times/sec with intervals of 0,01g. When balance 1 is ready to deliver a result from the item on the scale, the figures gained from balance 0, is delivered appx 1-2/60 of a second later. I would like to ignore the small time delay in gathering the result from the two balances. Based on these informations is it then possible to calculate the actual weight on balance 1 - based on a fomular?

7. If I am interpreting this correctly, you want to be able to calculate the mass of an object given its weight and the weight of a known mass.

I think the formula you are looking for is M_{object} = M_{known} \frac {W_{object}} {W_{known}} where W is the weight as measured by the scale and M is the mass measured on a stationary scale.

I don't think that the 1/60th of a second delay will throw this formula out significantly, but this depends on how accurate you want your scale to be. The best way to find out is to experiment with a known mass.