# compensate weight due to acceleration (loadcells)

I'm trying to make a weighing scale that works in a lift. I'm using load cells to measure the weight, let's say it's F. Then I use that weight F and g which i know is 9,81 m/s^2 to calculate the mass of the object like this.

``````F = m*g <=> m = F/g
``````

This works when the lift is in rest (or when moving with constant speed) but when it accelerates the weight F increases and because of that the mass also increases. Is there a way to compensate this weight(/extra force) due to acceleration?

• Maybe measure the effective g with an accelerometer? – Edgar Bonet Nov 14 '20 at 18:04
• Of cause it's possible. g is an acceleration. If the lift accelerates, you have to measure the acceleration, with an accelerometer. Then use the found acceleration instead of g in the formula. The resulting acceleration, you measure, is g plus(up)/minus(down) the lifts accelleration. – Peter Paul Kiefer Nov 14 '20 at 18:08
• 1. only weight when the lift isn't moving. 2. Have a separate load cell with a known weight on top, so you can calculate the acceleration. 3. Average the values over time, to average out the error due to the short moment of acceleration (most lifts only accelerate/decelerate for a few seconds). 4. Ignore it all together as lift only accelerate at 1m/s², giving you, at most, a 10% error is weight, which might be acceptable depending on the application. – Gerben Nov 15 '20 at 14:14

The force due to acceleration is: F = m*a

So to calculate the weight while accelerating the formula is:

F = mg + ma or F = m(g + a)

Note, of course, that a can be positive or negative depending on whether the elevator is accelerating or decelerating.

The trick here is to know what "a" is. The best approach is probably to get an accelerometer and use its reading in your calculations.

By the way, the mass is always a constant here. It does not change, in Newtonian Physics at least, due to acceleration.

• Note that an accelerometer actually measures a+g. – Edgar Bonet Nov 14 '20 at 18:11

You could measure the total acceleration (g + lift acceleration) with an accelerometer, and use that value in your calculation instead of 9.81m/s2. If the measured value is 0, sound an alarm.

Or, assuming the elevator only accelerates/decelerates for short periodes only (say, less than 2s) during a "trip", you could make a series of measurements in a loop at, say, every 0.5s.

When you get, say, 5 measurements that are the same, or very close to each other, you know acceleration = 0 and you can average those 5 measurements to get the desired result.

Values may vary depending on the lift, and the measurements will be slow in coming.