Elementary level question on Accelerometer scale factor

[ArixII]

Hi,
I've received a new accelerometer with a scale factor declared in datasheet as: 800 mV/g @ 1.5g. I googled and searched the forum but still can't fully be sure I understand the accelerometer and its functioning, so please correct me where I'm wrong:

One thing that very helped me to get some understanding, was this post:
http://tom.pycke.be/mav/69/accelerometer-to-attitude
where you see this:

giving the calculation of:
accelerometer = cos (theta) * gravity
theta = acos (accelerometer / gravity)

Formula is easy and beautiful, and this is my first question please:
what 'gravity' means there? Suppose the robot owning this accelerometer is on Earth surface, then this 'gravity' means 9.8m/s^2?

Next how to relate this to my accelerometer board output please?
I apply a 5V to its board and it has a 10bit ADC, so if I receive:
x(LSB) * 5(V)/1024(LSB)
now to get rid of V dimension I use the scale factor:
x(LSB) * 5(V)/1024(LSB) * (1/{800mV/g})
get rid of 'm':
x(LSB) * 5(V)/1024(LSB) * (1000/{800V/g})
or
x(LSB) * 5(V)/1024(LSB) * (10/{8V/g})
finally:
x*(25/4096)g

questions:
1. but this finall scale factor is a very small number! seems wrong :(
2. in data sheet the scale factor is declared: 800 mV/g @ 1.5g
how to take to account the 1.5g? I can imagine when the vertical upward is 1g and vertical downward is -1g, but can't imagine 1.5g.
3. Am I right with the g unit there? As long as I know, this is an embedded number like 9.8m/s^2 on Earth surface. So finally, having 'g' in both accelerometer and gravity in this formula:
theta = acos (accelerometer / gravity)
we remain with a scalar number and everything stands good at the end of the story, correct?!

Please help to make this clear!

 

[berkeman]

Hi,
I've received a new accelerometer with a scale factor declared in datasheet as: 800 mV/g @ 1.5g. I googled and searched the forum but still can't fully be sure I understand the accelerometer and its functioning, so please correct me where I'm wrong:

One thing that very helped me to get some understanding, was this post:
http://tom.pycke.be/mav/69/accelerometer-to-attitude
where you see this:

giving the calculation of:
accelerometer = cos (theta) * gravity
theta = acos (accelerometer / gravity)

Formula is easy and beautiful, and this is my first question please:
what 'gravity' means there? Suppose the robot owning this accelerometer is on Earth surface, then this 'gravity' means 9.8m/s^2?

Next how to relate this to my accelerometer board output please?
I apply a 5V to its board and it has a 10bit ADC, so if I receive:
x(LSB) * 5(V)/1024(LSB)
now to get rid of V dimension I use the scale factor:
x(LSB) * 5(V)/1024(LSB) * (1/{800mV/g})
get rid of 'm':
x(LSB) * 5(V)/1024(LSB) * (1000/{800V/g})
or
x(LSB) * 5(V)/1024(LSB) * (10/{8V/g})
finally:
x*(25/4096)g

questions:
1. but this finall scale factor is a very small number! seems wrong :(
2. in data sheet the scale factor is declared: 800 mV/g @ 1.5g
how to take to account the 1.5g? I can imagine when the vertical upward is 1g and vertical downward is -1g, but can't imagine 1.5g.
3. Am I right with the g unit there? As long as I know, this is an embedded number like 9.8m/s^2 on Earth surface. So finally, having 'g' in both accelerometer and gravity in this formula:
theta = acos (accelerometer / gravity)
we remain with a scalar number and everything stands good at the end of the story, correct?!

Please help to make this clear!

Yes, what they have called "gravity" would more accurately be called "acceleration due to gravity", or 1g = 9.8m/s^2.

So when you power this accelerometer and just hold it vertically, does it put out about 800mV? It is only experiencing 1g when you hold it still, so depending on what its calibration curve looks like, you might get something different (hopefully not too different). If you then turn it upside down, do you get -800mV? Or is it uni-directional?

What are its power supplies? Does it use split +/- symmetrical power supplies to give you bidirectional acceleration data? Or does it use a single supply, and have an offset in voltage for 0g?

Can you post a link to its datasheet?

 

[vk6kro]

The output is 0.8 volt for 1 G, so it would take (5 / 0.8), or 6.25 G, to give 5 volts out.

The A to D converter divides this 6.25 G into 1024 parts, so each is equivalent to 0.00610 G.

eg if you get a reading of 234, that is equivalent to 1.428 G. (ie 0.00610 times 234).

If you like, you can convert this 1.428 G to meters/second^2 by multiplying by 9.8
The 1.5 G figure is a measurement point only.
You do need to get some data on how linear the device is. That is, if the G force doubles, does the output double?

 

[sophiecentaur]

You could always measure its sensitivity yourself. Hanging it on a nice big penduum, you can get predictable peak values of g, if you release it from various heights.- e.g. from 90 degrees gives you 3g (whatever the pendulum length).

 

[D H]

One thing that very helped me to get some understanding, was this post:
http://tom.pycke.be/mav/69/accelerometer-to-attitude
where you see this:

giving the calculation of:
accelerometer = cos (theta) * gravity
theta = acos (accelerometer / gravity)

One word: Yech. Two more words: Double yech.

Accelerometers do not sense gravity. Period.

A 3-axis accelerometer at rest on the surface of the Earth will register an acceleration of about 1g directed up, not down. A 3-axis accelerometer at the center of mass of an orbiting spacecraft (or anywhere in an orbiting spacecraft that is not rotating with respect to inertial) will register very close to zero. Accelerometers sense acceleration due to every force but gravity. In the case of the accelerometer at rest on the surface of the Earth, the dominant force is the normal force from the ground. This normal force has a magnitude of about 1g and is directed upwards. In the case of an accelerometer in an orbiting spacecraft , the non-gravitational forces include atmospheric drag for a vehicle in low Earth orbit and radiation pressure. These are very small forces, so the accelerometer will register very close to zero.

In the case of that airplane, the non-gravitational forces are atmospheric drag, lift, and thrust. That is what the accelerometer is measuring, not gravity.

 

[sophiecentaur]

Accelerometers do not sense gravity. Period.

So, if you turn one upside down, it won't notice?

 

[D H]

So, if you turn one upside down, it won't notice?

What exactly do you mean by that, sophie?

Turn an accelerometer that is resting on the ground over and it will still register 1g upwards. Turn it in its side and it will still register 1g upwards. Turn it in any direction and it will still register 1g upwards. How that translates into accelerations as measured by the accelerometer of course does depend on the orientation of the accelerometer's case frame.

The software behind the inertial navigation systems on airplanes, missiles, and spacecraft needs to know the orientation of the accelerometer case frame with respect to the vehicle structure. The software also needs to know the orientation of the vehicle's structure with respect to some inertial frame. One way to do this is to integrate the measurements from the vehicle's gyros over time. Two problems here. One obvious problem is that the integration has to start with a known state. A less obvious problem is that gyro measurements are noisy. The time integral of white noise is a random walk.

Ignoring this problem and assuming that the orientation of the accelerometer with respect to some inertial frame are known, those accelerometer readings can be used as the basis for determining the vehicle's inertial velocity and position by once again using numerical integration techniques. Here we have a big problem: Accelerometers don't sense gravity. All aircraft, missile, and spacecraft that have an onboard inertial navigation system also have software algorithms that compute gravitational acceleration. What is integrated is the vector sum of the computed gravitational acceleration plus the acceleration sensed by the accelerometers.

 

[sophiecentaur]

What I mean is that if an accelerometer 'knows' in which direction, relative to itself, that gravity is acting, hasn't it just got to be sensing it? Or are you using the word "sense" in a different sense :grin:
Or, perhaps, there are accelerometers and 'accelerometers' and yours is different from mine. The one in my ipod certainly 'senses' g, cos it tells me when it's tipped and I'm sure it hasn't got a gyro inside it.

 

[D H]

The accelerometer is sensing everything but gravity, sophie.

Think of it this way: Suppose accelerometers did sense acceleration due to gravitation as well as accelerations due to other real non-gravitational forces. Were that the case an accelerometer at rest on the surface of the Earth would register a very small acceleration. That isn't what happens. They instead register an acceleration of 1g pointing up. They are measuring the normal force but not the gravitational force.

 

[sophiecentaur]

I assume we are in a spherical, non rotating Earth etc. for all of this.
If the accelerometer is stationary, the mass on the spring is in a non-equilibrium position, having moved 'downwards', it will register an 'up' signal. The only thing that has made it move is the presence of g. Without g, it would show a value of zero. The force (mg) which it registers is, of course, vertical, but how is it different in value from the gravitational force? Are you saying that the accelerometer is only 'inferring' the gravitational force by measuring the normal force?
Are you just being pedantic or is there a real point to your replies to my initial statement. I really can't see where you're going. The daft thing is that we both know the same theory so where are we parting company? Am I not allowed to invoke the equivalence principle?

 

[D H]

It is the equivalence principle that says you cannot sense gravitation.

Suppose you go bungee jumping and attach some accelerometers to your body. When you first jump off the bridge the accelerometers will register zero acceleration. Yet you are accelerating Earthward at about 9.81 m/s². The ~3×10^-6 m/s²/m variation in g with altitude certainly does not explain this reading of zero. So what gives? Simple: Accelerometers don't sense gravitation.

 

[sophiecentaur]

I understand what you are saying but you may as well say the same about an Ohmmeter not 'sensing' Ohms. It infers resistance by measuring Volts and Current or, alternatively, by comparison with the resistance of another, standard, resistance.
In fact, there are very few measuring / sensing devices that directly sense the quantity that is marked on the scale.
In the case of an accelerometer, we have two pieces of information which yield the value of g just as validly. We measure the weight of our suspended mass and we establish that the laboratory bench has zero acceleration. That gives us g.
At the start of your bungee jump there is no force on the suspension spring so we can't use the instrument to measure g.
The equivalence principle says that you can't tell the difference between acceleration and weight. But, if you know the acceleration of your reference bench, you can infer the value of g. How is that different from a thousand other measurements we do?

 

[D H]

The accelerometer is attached to you, sophie.

You are trying to invoke the equivalence principle and then go back to Newtonian mechanics. You can't do that because the equivalence principle does not hold in Newtonian mechanics. If you want to invoke the equivalence principle you must use all of general relativity, and gravitation is a fictitious force in general relativity. It isn't a real force.

 

[TurtleMeister]

In the case of that airplane, the non-gravitational forces are atmospheric drag, lift, and thrust. That is what the accelerometer is measuring, not gravity.

Yes.

Here's a video demonstration.

 

[sophiecentaur]

The accelerometer is attached to you, sophie.

You are trying to invoke the equivalence principle and then go back to Newtonian mechanics. You can't do that because the equivalence principle does not hold in Newtonian mechanics. If you want to invoke the equivalence principle you must use all of general relativity, and gravitation is a fictitious force in general relativity. It isn't a real force.

OK, don't use the equivalence principle - use the fact that g is the same, numerically, whether you are talking ms^-1 or Nkg^-1. An accelerometer works because of a force on a spring, producing a measured displacement.

Put the accelerometer on a different planet and it will register a different value of g, because the little mass in the middle will weigh less. So it is capable of measuring g. How can it measure it without sensing it? Moreover, it will tell you which is up and which is down. Of course it relies on an upward force from the bench and that bench needs to have a 'known' acceleration but many other measuring / sensing instruments need to input two quantities in order to 'sense' a quantity. What is so different about an accelerometer?

In the OP, the accelerometer will register g (vertically) in level flight. If it were calibrated to read zero in level flight then, when flying upside down, it would register -2g: very confusing for the navigator . Are you just saying that registering and sensing are two different things?

 

[D H]

OK, don't use the equivalence principle - use the fact that g is the same, numerically, whether you are talking ms^-1 or Nkg^-1.

The same as what?

Put the accelerometer on a different planet and it will register a different value of g, because the little mass in the middle will weigh less. So it is capable of measuring g.

No, it is not. It is measuring apparent weight, not actual weight (m*g). The apparent weight of some object is the vector sum of all real non-gravitational forces acting on the object (Newtonian mechanics point of view) or the sum of vector sum of all real forces, period, acting on the object (general relativistic point of view). From either point of view, Newtonian or general relativistic, accelerometers do not measure gravitation force. Actual weight can be computed in the special case that the relationship between apparent weight and actual weight is calculable. In the case of Einstein's elevator, the relation between actual and apparent weight is not known.

In the OP, the accelerometer will register g (vertically) in level flight.

it will register local vertical because that is tautologically the definition of local vertical. It will not register g (if by that you mean the acceleration due to gravitation), nor will it indicate the direction of the gravitational acceleration vector. Straight and level flight is not a straight line. It is a curved path that follows the curvature of the Earth.

If it were calibrated to read zero in level flight then, when flying upside down, it would register -2g: very confusing for the navigator .

Even in planes equipped with inertial navigation systems, the accelerometer readings typically are not presented to the pilot. There aren't very informative (to the pilot).

Are you just saying that registering and sensing are two different things?

No. I am saying that accelerometers don't measure gravitation.Three questions:
1. The gravitational acceleration of a spacecraft in low Earth orbit toward the Earth is about 0.9 g. So why does the spacecraft 's accelerometer register an acceleration that is very close to zero?

2. Why do aircraft, missiles and spacecraft outfitted with an inertial navigation system invariably compute gravitational acceleration via some software model?

3. For an object at rest on the surface of the Earth, why are apparent and actual weight equal in magnitude only at the north and south pole?

 

[sophiecentaur]

"The same as what?" ---- read the whole sentence.You have missed my point entirely. The three questions at the end have obvious answers and they are all to do with the fact that, to 'measure gravity' you need to know more than just what the accelerometer tells you.

I could turn this around a bit and ask you how G (big g) can be measured by anything other than, effectively measuring the force on a mass (two masses) and a spring. (not a million miles from an accelerometer design)

What is so special about gravitation that makes you say that measuring it must involve more than its effect alone? You need to allow for other factors / motions (aeroplane's flight or Earth's rotation) in the same way that you need to include the resistance of connecting wires (etc. etc.) when measuring resistance accurately. In fact, electrical meters ofter work by inferring V from I (moving coil meters) or I from V (DVMs) so it must be permitted to get a value of something by 'inference' and it need not be direct measurement.

You seem to have some special message about this thing which is no more than to say that g measurements need to have other things factored in.
An accelerometer will tell you just the force on the little spring. That force is due to g and some other things. But you did originally say that an accelerometer doesn't "sense gravity. Period". I don't know what else the word "sense" can mean if it doesn't mean that its readings change when g changes and that, once other factors are accounted for, it will give a value for g. Does any other form of 'anythingmeter' do anything more fundamental than that?

All this sound like the sort of conversation I used to hear, as a lad, in School when Physics teachers forbad you to use the expression 'centrifugal force' to describe what happens to a whirling conker. In a desperate attempt to dissuade students that things fly out radially, when released they forgot to acknowledge that things do, actually, fly further away from the place to which the string was attached and that there is a force on the string which is away from the centre of rotation.

 

[D H]

"The same as what?" ---- read the whole sentence.

So are you just saying that N/kg is the same as m/s²? If so what is the point?

What is so special about gravitation that makes you say that measuring it must involve more than its effect alone?

What is so special about gravitation from the perspective of Newtonian mechanics is that there is no such thing as a gravity shield. What is so special about gravitation from the perspective of general relativity is that gravitation is not a real force. It is a fictitious force. You can't measure fictitious forces with sensors.

 

[sophiecentaur]

So are you just saying that N/kg is the same as m/s²? If so what is the point?

Just that, as you go from planet to planet, the values will still track and the effects of gravity and acceleration will be indistinguishable.
You seem to be taking it as far as saying that an accelerometer cannot be aware of the effects of gravity. I know that you can't mean that.

What is so special about gravitation from the perspective of Newtonian mechanics is that there is no such thing as a gravity shield. What is so special about gravitation from the perspective of general relativity is that gravitation is not a real force. It is a fictitious force. You can't measure fictitious forces with sensors.

You can call gravity a force or anything else but you can't deny that, where there is no gravity and nothing is being accelerated by any other means, an accelerometer will register zero. Introduce one or the other and it will start to register- possibly with the help of a planet to sit it on, but that's all part of the measurement equipment.

What has a (non) gravity shield got to do with it?

Did you read what I wrote about centrifugal force? Do you not see a parallel?

 

[D H]

You seem to be taking it as far as saying that an accelerometer cannot be aware of the effects of gravity. I know that you can't mean that.

That is exactly what I mean. That accelerometers cannot be aware of the effects of gravity is precisely why inertial navigation systems have to compute gravitational acceleration using Newton's law.

Another example: Suppose you strap an accelerometer on your chest, strap a parachute on your back, and go skydiving from a hovering helicopter. Just before you jump off the accelerometer will read 1g up. Just after you jump it will read near zero. Gravitational acceleration hasn't changed at all and yet the accelerometer reading changed by a huge amount. During free fall the accelerometer reading will gradually increase, reaching 1g when you reach terminal velocity. The accelerometer reading will spike when you pull the rip cord and then quickly fall back to 1g as you reach a new reduced terminal velocity. The reading will spike once again when you hit the ground. That accelerometers sense everything but gravitation readily explains all of these variations.

Did you read what I wrote about centrifugal force? Do you not see a parallel?

Yes. The parallel is that you should have listened to your instructor.

All this sound like the sort of conversation I used to hear, as a lad, in School when Physics teachers forbad you to use the expression 'centrifugal force' to describe what happens to a whirling conker. In a desperate attempt to dissuade students that things fly out radially, when released they forgot to acknowledge that things do, actually, fly further away from the place to which the string was attached and that there is a force on the string which is away from the centre of rotation.

Let's look at what an inertial observer sees. The "whirling clonker" does not fly out radially. It flies out tangentially. To this inertial observer there is absolutely no difference in behavior between a "whirling clonker" and one that is tossed. None. The inertial observer has no reason to invoke this fictitious centrifugal force.

This fictitious force only is needed when you try to explain the motion of the whirling clonker from the perspective of a rotating observer, and then only if you are trying to explain the observed motion in terms of Newton's second law. Now suppose instead of releasing this "whirling clonker" you are spinning around and watch someone toss a clonker with exactly the same velocity as that given to the whirling clonker. The tossed clonker exhibits exactly the same behavior.

Let's take this a bit further afield. The stars appear to be undergoing a huge acceleration. Is Alpha Centauri really accelerating Earthward 22,000 g, or is this apparent acceleration just a fiction of our Earth-bound perspective?

 

[sophiecentaur]

I'm afraid I have to be more pragmatic than you, DH.
If I have a meter that tells me the value of g on the surface of the Earth (despite some corrections that need to be made) then,as far as I'm concerned, it is measuring g. It may not work in some conditions -such as free fall - but why should that bother me when I want to know what g is where I am standing?

If I have a mercury in glass thermometer, I can believe the temperature written on the scale because, somewhere down the line, it was calibrated against a 'better' standard which would work over a much bigger range of conditions than my device. It still 'senses' temperature, even though it's modus is different from the reference instrument. Does that, somehow, make the temperature I read invalid?

And. as for listening to my tutors, that's just what I did and then I listened to more informed tutors and reconciled the limited view I got at School with a wider view. Even in a stationary reference frame, a conker moves visibly away from the original central attractor - but not radially. That's still outwards, isn't it? Red Blood cells find themselves at the outside of a centrifuge - which is further away from the centre than where they started. Forget about frames of reference, there's a force (or what else?), causing them to move within the plasma. It really doesn't help to get up oneself too much about these matters or you can't see the wood for the trees.

Re Alpha centauri - my conker may have an acceleration towards my hand whilst I am whirling it but the string isn't getting any shorter. Are you saying that the star is accelerating towards us at that rate? I'd like to read more about that. Have you a reference, please?

 

[javiramos2008]

Bahhhhh. You CAN measure the gravity vector with an accelerometer. Of course, the accelerometer is a tiny mass mounted on flexures. If you rotate it around, the gravity vector change wrt to the case of the accelerometer. This is assuming that the accelerometer measurements are taken in a static condition. In other words, you CAN use an accelerometer to measure the tilt of a static device (I do that in my lab). And indeed, it is a very cheap way to do it.

Buuuut, an accelerometer will measure the sum of the forces acting on it. So as P H said, if the accelerometer is free falling it will sense 0. If it is accelerating then it will measure some other thing of course.

waduppp physics?! I am an engineer so forget all the theo talk about Einstein etc. Good for a coffee talk, but let's be practical here please.

 

[D H]

Bahhhhh. You CAN measure the gravity vector with an accelerometer. ... This is assuming that the accelerometer measurements are taken in a static condition.

You are not measuring gravity here. Think of it this way: An accelerometer that is sitting still on the top of a table will register an acceleration of about 1g directed upward.. The acceleration due to gravitation points downward, not upward. What you are measuring is the acceleration due to all of the forces acting on the accelerometer except for gravity. It is the table (contact force) and the atmosphere (buoyancy) that are pushing the accelerometer upward -- and that is what the accelerometer is sensing.

This sensed upward acceleration can be used as a proxy for gravitation with the assumption that the object is stationary with respect to the rotating Earth -- but you are not measuring gravity. Attach that accelerometer to an apple and let the apple fall and the accelerometer will read zero.

This distinction might not be so important for things that are essentially stationary with respect to the rotating Earth. Simply make the stationarity assumption, filter out high-frequency changes to address minor violations of that assumption, and voila! you have a proxy for gravity. The distinction is extremely important for inertial navigation systems on aircraft and spacecraft . The navigation software needs to augment the sensed accelerations with a computed gravitational acceleration to get anywhere close to an accurate representation of the vehicle's state over time because gravitation itself is unmeasurable.

Even with the stationarity assumption, the sensed acceleration is not yet a proxy for gravitation. The stationarity assumption pertains to the rotating Earth, so you are implicitly dealing with a rotating frame of reference here. You have a proxy for gravity, not gravitation (see [post=1602624]this post[/post]). The distinction in physical geodesy between gravity and gravitation is small given our small rotation rate of 1 rotation per sidereal day. The distinction would be much greater if a day was only 90 minutes long.

 

[sophiecentaur]

@javiramos
You can't win here - I have made all these objections already and they are 'reasonable' but rejected. I have become reconciled with what the purists are saying but I agree with you that you CAN measure g with an accelerometer AND a stationary mounting. You need to be able verify that the mounting (table top) is not accelerating, though. It could be slowly sinking into the ground or rising and then you'd get the wrong answer. AS an Engineer, at heart, I'd go for that measurement, though.

 

[D H]

I have become reconciled with what the purists are saying but I agree with you that you CAN measure g with an accelerometer AND a stationary mounting.

Two comments:
1. You are not measuring g. You are measuring a proxy for g based on the assumption that the object is not accelerating with respect to the rotating Earth. Slap that accelerometer on an apple up in a tree. The accelerometer will measure approximately 1g upwards (not downwards) up until the moment the apple breaks loose from the tree. It will then register zero. This drastic change in the measured value doesn't mean gravity has changed. The gravitational force on the apple hasn't changed one iota.

2. Even when the proxy measurement is valid (i.e., the object is indeed stationary with respect to the rotating Earth), you still are not measuring the gravitational acceleration. The proxy measurement is the vector sum of the gravitational acceleration plus centrifugal acceleration. That centrifugal acceleration is small (about 1 part per 500 at the equator), but it is there. Even a lousy off-the-shelf MIMS accelerometer has much better than 3 digit accuracy.

 

[sophiecentaur]

Two comments:
1. You are not measuring g. You are measuring a proxy for g based on the assumption that the object is not accelerating with respect to the rotating Earth. Slap that accelerometer on an apple up in a tree. The accelerometer will measure approximately 1g upwards (not downwards) up until the moment the apple breaks loose from the tree. It will then register zero. This drastic change in the measured value doesn't mean gravity has changed. The gravitational force on the apple hasn't changed one iota.

That's precisely what I said: "you CAN measure g with an accelerometer AND a stationary mounting"
If you like, you can do the measurement at one of the Poles or, using some extra information about radius and rotation, then you can calculate g.
What amazes me is how this is, somehow, regarded as 'special' as measurements go. If you want to measure almost any physical quantity you need to make multiple measurements of a number of different quantities. All that stuff about falling apples is totally obvious and, in the same way as you'd do an electrical measurement or an energy measurement, you would use extra information to get your answer.