For the last three days, I have been passing a kidney stone, something I have become all too accustomed to over the years. An undeniable pattern has begun to emerge. Whenever I fly, there is a high probability I will start passing a stone within 48 hours. Such has been the case on 3 out of my last 4 flights. I decided to try and figure out why this might be so. My first theory was that it had something to do with the change in gravitational force at high altitude. When I started researching the subject, I discovered that this was just the tip of the iceberg. Below, I have copied excerpts of a Quora posting by avid traveler and airline pilot Bruno Gilissen. I found it absolutely fascinating but also concluded that these effects were not sufficient in magnitude to trigger kidney stone passage. Later in the week, I will investigate the effects of cabin pressure change for a possible clue.
Assuming you know already that there’s a difference between mass (amount of matter) and weight (the force of gravitational pull), we’ll assume that your mass would remain constant during the measurements. Of course, in reality, anything you consume makes you gain mass, anything you do in the toilet makes you lose it, and sweating and breathing make you lose mass too.
There are two ways to weigh yourself then:
- by using a mechanism of balancing weights
- by using a mechanism of resistance, usually a spring
(When using method 1, in different environmental conditions, whatever happens on one side of the scale also tends to happen on the other side. With method 2 you’re relying on a spring that is calibrated for a fixed set of conditions and those conditions aren’t necessarily the same everywhere.)
Let’s see what forces these scales will be measuring, and deduct how they each will be affected.
I. Gravity, obviously
As per Newton’s law of gravity, applied to our planet:
𝐹𝑔=𝐺𝑀𝑒𝑚2𝑟2Fg=GMem2r2
For both the conditions on the ground and on an airplane, the mass of the earth (𝑀𝑒Me), and the gravitational constant (𝐺G) remain the same. We also assume your mass (𝑚2m2) stays the same. So the only change here depends on the distance between the center of mass of earth and yourself, squared (𝑟2r2).
That distance 𝑟r would entirely depend on where the airplane is in the world and on the height above the earth.
(Earth bulges around the equator and is flattened a bit at the poles. The 6371 km Google will reveal as our planet’s radius, is only based on that formula in red.)
When you’re flying a polar route and you’re close to the north-pole, you’ll be closer to earth’s center of mass than when you’re flying near the equator.
Checking out how much this matters:
(Comparing different scenarios, you’ll notice that our weight will stay well within a percent difference, no matter where on earth or how high the airplane is. This table is normalized for someone who weighed himself at sea level at the equator first.)
(A table normalized for people who weighed themselves at a more average latitude at sea level first. You’ll start weighing more at the poles, less at the equator, and less the higher you fly.)
In an average scenario where you’re in an airliner cruising around 10 km high, you can perhaps lose up to half a percent of weight due to gravity. That may become even up to a full percent if you’ve first weighed yourself somewhere far up north like in Barrow (Alaska, USA) or Thule (Greenland) and you’re flying now very high in a private jet over Ecuador. (Private jets tend to cruise higher than airliners.)
Now, obviously, if you stand on a scale that’s based on a counterweight, the weights doing the countering are subject to the same reduction or gain you’re getting on the one side. So that gets you nowhere. The difference then is zero.
But if you took the home scale from in your bathroom, and check the difference, then you’ll notice you perhaps lost or gained a few hundred grams. Someone who’s 70 kg can realistically lose up to 350 grams. If you want to lose a tad more, then you need to get on a private jet.
II. Apparent forces because of movement and frame of reference (even when flying straight and level):
II.A. First the easy one: the movement of the airplane itself, in relation to its immediate environment.
You’ll know from that funny feeling you get in an old elevator that your apparent weight can change due to an acceleration. After all, when weights on a scale are accelerated upwards or downwards, that acceleration converts to a force that either pushes more or less on the scale. We don’t need Einstein’s general relativity to understand that, I hope, but if you doubt it I suggest you try it next time you take an elevator.
The home scale you stand on in the plane may similarly reflect the accelerations of the ship. You’ll understand that the movements of the needle during turbulence will swing further than the half percent from the previous example. For a scale based on balancing weights there will also be sideways accelerations in the horizontal plane that skew measurements (by changing the arm), even though theoretically purely vertical accelerations should have no effect on this device.
Not only turbulence will affect this. The pilots or the autopilot will continuously make small inputs that may cause your scales to deflect too. Or large inputs, of course, when turning or changing level for example. To take it to an extreme example: check out parabolic flight, which they use with astronauts to simulate weightlessness.
II.B. The rotation and frame of reference of the earth: coriolis effect.
Because the earth rotates, and we therefore use a rotating big ball in space as if it’s our own frame of reference, some apparent forces exist; apparent forces that can also influence our measured weight.
(At the pole you’d only turn around your own axis in 24 h while standing still, while at the equator you’ll have followed the path of a giant circle of more than 40,000 km in that time. The researcher at the pole doesn’t move while the explorer at the equator, unconscious of it, travels bloody fast to do all that distance in that time. That has consequences.)
The effect of a rotating earth is that for us on earth, it seems to bend things off. Seen from far away in space, those things don’t violate Newton’s laws: they still want to continue their paths straight and at constant speed unless acted upon by a force. But if you stand on that rotating circle, what’s straight for an astronaut’s point of view is not going to remain straight for you; and similarly your straight path will seem curved from Major Tom’s point of view.
(What seems to move logically in a straight line for the astronaut, has to look bent-off on earth.)
Unfortunately this effect, called coriolis effect (some physics books dislike the term “force” because it’s not a basic force of nature but a consequence of frame-of-reference-changes), is often misunderstood. Most people think it only works in a horizontal sense because they’ve learned that it happens when you travel from a place on our surface where earth rotates faster to where it turns around slower, or vice versa.
But there’s a vertical aspect too. If you move 10 km up into the air, you are also increasing your distance to earth’s center by 10 km, so if you don’t “catch up” with the surface, you’ll be 10 km behind at the end of the day (times 2 pi, times the cosine of your latitude, to be correcter). That means there’s an apparent force that bends you off against the direction of earth’s rotation if you climb. Similarly, if you descend you’re going to have to adjust speed and course ever so slightly to not miss the point you were aiming for, because you’re reducing the radius to earth’s center.
(The plane that took off flies further away from the center of rotation after it climbed, so it needs to travel a bit further now to keep up. It looks as if there’s a mysterious force that makes it want to stay behind, compared to earth’s surface.)
The coriolis effect exerts this apparent force on the scales:
𝐹𝑐→=2𝑚Ω⃗ ×𝑣𝑎→Fc→=2mΩ→×va→
For an airliner cruising at 480 knots groundspeed (𝑣𝑎va), that translates to something like maximum 0.04 𝑚/𝑠2m/s2. It would depend on the direction of travel and the latitude (cross-product of vectors, whereby Ω⃗ Ω→ is earth’s rate of rotation; and – if you want to calculate it – consider a sidereal day). Flying at the equator that acceleration is purely vertical, with no horizontal component, and opposing the weight: that makes you lighter.
On the balance-style scale this would not matter. But on the scale with the calibrated spring this will affect the measurement. If it is calibrated for an average gravitational acceleration of 𝑔=9.80𝑚/𝑠2g=9.80m/s2, then the 0.04 𝑚/𝑠2m/s2 is just over 0.4% of that. The largest difference is if you weigh yourself first at the surface and then compare it to what’s on that scale at the moment you overfly the equator.
(Difference-table for someone who weighed himself at rest on earth first. The faster you fly, and the closer to the equator you are, the bigger the difference.)
Taking again the example of someone who’s 70 kg, the scale may show a weight loss of less than 300 grams thanks to coriolis.
II.C. The curved path around earth of the airplane: centrifugal and Euler-force
You have to counter the fact that you follow earth’s curvature too. That means that from the astronaut’s point of view, you are constantly accelerating a smidgen downwards (your downwards, on earth). That makes you, flying curved on earth, feel a fictitious centrifugal force, which has a vertical component that opposes your weight.
(Because you follow a curved path there is an acceleration required to keep it curved. From the perspective of the one in a vehicle that makes this curve, there’s an apparent force making him fly outwards… like when you turn in a car and you slide to the outside of the curve.)
The magnitude of that centrifugal “effect” can be calculated with this formula:
𝐹𝑓=𝑚𝜔2𝑟Ff=mω2r
Where 𝜔ω and 𝑟r depend on the latitude and course of the airplane.
The maximum possible values for all the factors will happen when both turn radius and angular velocity are greatest. For earth that means for example flying perfectly east at the equator. Therefore, you could feel an acceleration upwards up to (2𝜋𝑇)2𝑟(2πT)2r = 0.03 𝑚𝑠2ms2. That’s for a flight at 14 km or almost 43,000 ft high, the ceiling of an Airbus A350.
Again, compared to a 𝑔g of 9.80 𝑚/𝑠2m/s2 that’s about 0.3%. For our volunteer of 70 kg, it may make him up to 240 g or so lighter… provided he uses the bathroom scale, not the balance.
Additionally, there’s also an ever so slight apparent force when 𝜔ω changes, for example by suddenly altering course and/or latitude. But that is really so small that it’s not worth considering. The force is called Euler force (or “-effect”, contingent on who wrote the book) and can be calculated with this formula:
𝐹𝐸=𝑚𝑑Ω𝑑𝑡×𝑟FE=mdΩdt×r
where 𝑑Ω𝑑𝑡dΩdt is the change of of angular speed around the point considered. Since we’re not really going to change that factor anywhere near significant, forget that. It would only come into play on “the day the earth stood still” or possibly when doing a very quick course reversal in equatorial regions. Unrealistic.
II.D. Other accelerations you forgot about: Eötvös effect
From the astronaut’s viewpoint, it also matters whether you’re flying east or west.
Suppose that you are over the very north of Canada, flying perfectly west at a speed just countering earth’s rotation. That would look like you’re stationary to the astronaut.
(Our astronaut would see the red airplane 1 as if it remains stationary while earth rotates under it. He would see the purple airplane 2 fly with a very fast rotational speed.)
Important side note! Please please please people, do not confuse this with the ludicrous idea that you can just take off, hover in the air stationary, and arrive somewhere west of that without using energy! I cannot stress enough how often I’ve heard that silliness. Attempting to do that will require lots of fuel, since we need – from the astronaut’s point of view – to “stop” the motion the earth gave us already in the first place. From our perspective on earth, you are actively flying west. Don’t forget you’re changing frame of reference here! The laws that govern our universe still apply. Walk upstairs on a downward moving escalator for a whole day so that it looks you remain stationary to your friend sitting in the bar nearby, to test if that was energy-free, if you don’t believe me.
But back to the point.
When the astronaut would see you in a for him stable location there, your angular velocity 𝜔ω is zero. On the other hand, when you fly east at the equator, your velocity has to be added to the earth’s and your 𝜔ω increases. You’ll speed up for the astronaut, even though for you on earth it may look equivalent.
You hopefully see that it follows then that the velocity in your east-west direction matters. Also, don’t confuse it with the Euler force mentioned before, which is because of a change in angular velocity (angular acceleration) 𝑑Ω𝑑𝑡dΩdt. Here we’re talking not about a sudden change, but a stable velocity 𝑢u, which is purely the east-west speed and direction.
The formula for the acceleration due to Eötvös effect is:
𝑎𝑢=2Ω𝑢cos(𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒)+(𝑢2+𝑣2)𝑟au=2Ωucos(latitude)+(u2+v2)r
where ΩΩ is the rotation rate of the Earth, 𝑢u is the east-west velocity, 𝑣v is the north-south velocity, and 𝑟r is the radius.
The consequence is that westward-travelling people will feel heavier due to a downward acceleration, while flying eastbound gets them deflected upwards and thus measure lighter. But only on the spring-powered scale again. The balance-scale, once more, will not be affected because it happens on both sides of the balance-point equally.
(The change in indication on the bathroom-style scale if stationary at the equator is the reference. Poles aren’t calculated because there’s no east-west there; at the north pole everything lies south.)
So our subject of 70 kg, having left Quito in Ecuador and now flying full cruise speed towards Belém (Brazil) in the east, will feel some 300 g lighter. When he travels westbound at the same speed instead, he’ll feel 190 g heavier. It’s not symmetrical because earth’s rotation has to be taken into account, with westbound subtracted from it and eastbound added.
In effect, you are correcting your rotational speed with a term in this Eötvös effect: what you either counted too much or not enough when you only considered earth’s rotation. After all, for the astronaut your rotational speed is either more or less than the rate of rotation from our planet, if you’re moving east or west.
III. Archimedes force, or boyancy
Next, let’s have a look at Archimedes force. The issue here is that at higher altitudes the pressure differential between your feet and head (simplified) will not be the same than the differential at sea level.
(Something submerged in a fluid – or gas, like air – experiences a net upwards force due to the submersion. That’s because the pressure under the thing is higher than the pressure above it, so it gets pushed upwards. It’s why you and heavy ships float on water: density, or pressure, is high under you, while low above.)
Indeed, at altitude the pressure differential below and above you would be ever so different. That’s because the pressure of the air doesn’t evolve linearly with altitude, but quasi-logarithmic.
(If you go higher (right in the graph), then air pressure drops. But the difference in pressure per unit becomes less and less, as it’s not linear. Near the surface (left side of the graph) it falls quickly per altitude unit. High up (right side) it drops a lot slower.)
The misconception here may be that you’re exposed to those very low pressures. You’re not. You fly in a pressurized tube so you can still breathe. An airliner’s cabin pressure is usually maintained at something around a 6,000 up to 8,000 ft equivalent, following a programmed schedule that looks at the plane’s altitude and length of flight (which has to do with cycles of pressurizing it and structural fatigue).
So what you need to consider then, regarding Archimedes force, is the variation of pressure difference between your feet and head (simplified) at sea level and at say 7,000 ft; even though you may be actually cruising at 40,000 ft.
(The Archimedes force pushing someone up at sea level, versus at 7,000 ft cabin altitude (which is not real altitude you fly at!))
So there’s a difference of give or take 3 Newton, depending on how high our person is and how broad his shoulders and belly are, I guess.
3 Newtons is calibrated on the scales as 300 g, give or take. Again, we’re in the same order. Not all our cross section’s surface is of course at feet-level and head-level, so an educated guess of mine is that it would rather be half of that… but I stand corrected if you know better.
This time though, take note that the pressure differential will happen on your body, but it may not on the balance-style scale. It would depend on how large the counterweights are. Realistically they are not tall enough to matter, but if you are balanced by another person, then it may more or less even out again.
Therefore, realistically this difference will apply to all types of scales.