# Properly calculating G-factor

This may sound rather simple, but I was told that technically, G-force readout should actually indicate zero-G on the earth's surface, as well as in a stable orbit. This is due to the fact that G-force is measured using accelerometers (which measure a change in velocity) and since we down here on Earth surface, when standing still, don't change our velocity (neglecting earth's rotation effects), our G-factor should be 0. However, I'm pretty sure we here on Earth experience 1G. Now, I understand the physics of it: Our "fall" towards the center of the Earth is being blocked by ground, and we feel that as "Normal" force, or "weight". But how do you quantify and calculate this G-value in a unifying formula that works for any point in the universe?

And - let's keep things simple:

• Earth being a perfect, non-rotating sphere with uniform mass
• No influences of other solar bodies

All I want to know is how the G-factor is calculated using a force vector summation formula that can be applied to on-ground, in LEO and BEO scenarios.

• +1 A good answer to your excellent and challenging question will go into some depth, and point out both that you can't neglect Earth's rotation, gravity changes depending on altitude, latitude, what's underneath the ground locally, proximity to mountains, direction towards the Moon, and the Sun, etc, and that real-world accelerometers usually measure much more than "a change in velocity". – uhoh Apr 8 '19 at 6:30
• See for example @DavidHammen's table of related effects and his other answer and possibly this answer for example. To explore three ways down (or up) can be defined, see this answer. – uhoh Apr 8 '19 at 6:30
• Okay that's a cleaner question for sure. You should edit your question and modify it directly, rather than do that in comments because people don't always read through comments before writing answers. Thanks! – uhoh Apr 8 '19 at 22:35
• Great,I just updated it! Thanks! – Mitch99 Apr 9 '19 at 7:09
• Looks great, thanks! – uhoh Apr 9 '19 at 7:20

This may sound rather simple, but I was told that technically, G-force readout should actually indicate zero-G on the earth's surface, as well as in a stable orbit.

Whoever told you that was wrong. The G force reading should indicate zero g for a non-thrusting spacecraft in orbit well above the Earth's atmosphere, and also zero g the moment after a bungee jumper steps off a bridge. The G-force reading for a person standing still on the surface of the Earth watching the bungee jumper will be about 1 g, directed upward.

Suppose the spacecraft mentioned above starts thrusting in order to, for example, transfer to a higher orbit, or to go beyond Earth orbit. Its "G force" sensor will now register a non-zero value. The "force" in "G force" is a bit of a misnomer. G force has units of acceleration, not force, with 1 g being 9.80665 m/s2. What's common between the non-thrusting spacecraft and the bungee jumper is that the only force acting on them is gravitation. What's common between the thrusting spacecraft and the person standing still on the surface of the Earth is that some force in addition to gravitation acts on them. This suggests a better name for "G force": Net non-gravitational acceleration. That's a mouthful. An even better name is proper acceleration.

This is due to the fact that G-force is measured using accelerometers (which measure a change in velocity) ...

Accelerometers do not measure change in velocity. An accelerometer in a non-rotating, non-thrusting spacecraft above the Earth's atmosphere will register 0 g, even though the spacecraft's velocity vector is changing all the time. An accelerometer at rest on a table on the surface of the Earth will register 1 g directed upward. Smart phones use an accelerometer to determine which direction is up (or down), and this in turn is used to determine whether the cell phone should operate in landscape or portrait mode. Accelerometers instead sense (imperfectly) proper weight per unit mass -- i.e., proper acceleration, or "G force".

A perfect accelerometer would be the ideal device for measuring "G force". Note that I qualified what accelerometers measure with "imperfectly". Real accelerometers, as opposed to perfect ones, have a number of imperfections. A real accelerometer might register 1 g while a perfect one would register 1.005 g, or 2 g in a situation where a perfect accelerometer would register 2.01 g. This is called a scale factor error. Every reading is incorrect by a common factor. Another error is bias. For example A real accelerometer with a bias might register 0.005 g while a perfect one would register 0.0 g, or 1.005 g in a situation where a perfect accelerometer would register 1 g. Every reading is off by a constant amount. Yet another kind of error is noise. The readings from cheap accelerometers (e.g., the ones used in cell phones) are rather noisy. Even the very best cryogenically cooled, superconducting accelerometers exhibit some amount of noise.

All I want to know is how the G-factor is calculated using a force vector summation formula that can be applied to on-ground, in LEO and BEO scenarios.

That accelerometers do not measure acceleration due to gravity presents a challenge for spacecraft that self-navigate their position/velocity state. Such spacecraft need to have an onboard model of gravitation so that the acceleration due to gravity can be added to the accelerometer reading. This calculated value will be somewhat erroneous. The model of gravitation is never perfect (it's a model), and if the spacecraft's estimate of where it is in somewhat incorrect, the calculated gravitational acceleration vector will be incorrect. That accelerometers have errors that make their measurements of acceleration due to non-gravitational forces presents a challenge for spacecraft that self-navigate. The combination of the errors from computing the gravitational acceleration and the errors from the accelerometer means that the integrated acceleration (i.e., velocity) will drift from truth, and the doubly integrated acceleration (i.e., position) will do worse than drift from truth.

This means self-navigating spacecraft (and also self-navigating cars) need outside help to keep their estimated position/velocity state close to reality. GPS works great for vehicles in low Earth orbit and up to perhaps a bit beyond geostationary orbit. Beyond that, self-navigating position/velocity state gets much more challenging. A lot of spacecraft do not self-navigate their position/velocity state. For example, the New Horizons spacecraft that flew past Pluto and more recently Ultima Thule, had no clue where it was in space. Like many spacecraft, the onboard navigation only computed the vehicle's attitude/attitude rate state. New Horizons took pictures of Pluto and Ultima Thule via timed commands that told the vehicle times at which to change it's attitude/attitude rate and times at which to operate various instruments.

That accelerometers do not measure gravitational acceleration and that they imperfectly measure non-gravitational acceleration similarly presents challenges for self-driving cars on the surface of the Earth. The errors that inherently result from accelerometer-based dead reckoning would quickly make the deduced position and velocity worthless. Self-driving cars need outside help such as GPS (but GPS is rather lousy in cities) and maps (but maps are always out of date), and also nonlocal sensors (e.g., cameras).

• "The G-force reading for a person standing still on the surface of the Earth watching the bungee jumper will be about 1 g, directed upward", ->why upward? – Hobbes Apr 9 '19 at 14:20
• @Hobbes - Because "G-force sensors" (aka accelerometers) cannot sense gravity. (No local experiment can per Einstein's equivalence principle.) The Newtonian forces acting on a person standing still on the ground are the downward gravitational force and the upward normal force that keeps the person from sinking into the Earth. Accelerometers can only detect the latter, so 1 g upward. – David Hammen Apr 9 '19 at 14:46
• Another way to look at it: Accelerometers measure acceleration relative to a local stream of free-falling apples. To a person standing still on the surface of the Earth, that stream of free-falling apples is accelerating downward. Relative to one of the free-falling apples, the person is accelerating upward. – David Hammen Apr 9 '19 at 14:46

This may sound rather simple, but I was told that technically, G-force readout should actually indicate zero-G on the earth's surface, as well as in a stable orbit. This is due to the fact that G-force is measured using accelerometers

This is incorrect. G is not zero on Earth's surface. You're conflating two things: G-force, which is a force as a result of gravity, and acceleration, which produces a force as a result of movement.

This is due to the fact that G-force is measured using accelerometers

Accelerometers are a cheap way to measure acceleration. They cannot measure force, so they cannot be used to measure G due to gravity directly. You need e.g. a weighing scale to measure G.

Now, you can combine gravity and acceleration into a net force that acts on a body. This is just the addition of two vectors: the gravity vector, pointing at the center of the body that produces the gravitational force, and the acceleration vector produced by the vehicle you're in.