50 ms·

Does that mean that when you're flying an X15 you lose 6.2/22 = 28% of your bodyweight to the Coriolis force? That's pretty neat.

jacquesm 7y ago
That depends on your orientation.
- deleted 7y ago
  [deleted]
pdonis 7y ago
> Does that mean that when you're flying an X15 you lose 6.2/22 = 28% of your bodyweight to the Coriolis force? I think you mean centrifugal force. Unfortunately, no, you don't feel quite that much less weight, because centrifugal force goes like the square of the velocity, so the ratio is the square of 6.2/22, or about 0.08, i.e., only 8%.
- bzbarsky 7y ago
  There is a Coriolis force here too; it's just much smaller than the centrifugal force. Consider a plane flying at ground level along the equator at groundspeed velocity v in the direction of the Earth's rotation. In the Earth's (rotating) reference frame there is a centrifugal acceleration based on the Earth's angular velocity and a Coriolis acceleration given by -2 W x v, where W is the Earth's angular velocity and v is the plane's velocity. This ends up pointing straight up in this configuration. So in the plane's reference frame there is a straight-up centrifugal acceleration, given by (W + v/R)^2 * R = W^2 * R + 2W x v + v^2/R which is balanced by the aerodynamic and gravitational forces on the plane (which in this frame is non-accelerating), while in the Earth's reference frame there is a straight-up centrifugal acceleration given by W^2 * R and a straight-up Coriolis acceleration of 2W x v, and the plane is accelerating radially down at v^2/R due to gravity and lift combined. Now to put numbers to this, the speed record for an X15 is about 2000 m/s, R is 6.4 * 10^6 m, and W is 1 revolution per 24 hours, or 1/86400 s^1. So the three force components look like this, if I am doing my numbers right: W^2 * R is .00086 m/s^2 2W x v is 0.04 m/s^2 v^2/R is 0.63 m/s^2. So the v^2/R "centrifugal force due to the plane's motion" term is by far the biggest contributor here.
  - pdonis 7y ago
    > There is a Coriolis force here too; it's just much smaller than the centrifugal force. Yes, agreed.
mannykannot 7y ago
I think that, of the so-called inertial or ficticious forces, you are thinking of the centrifugal rather than the coriolis force [1]. On any of the ballistic / into space flights, you would experience weightlessness between the engine cutoff and the point where atmospheric forces became noticable -- this is what the space tourism companies are aiming for. I do not know if any flights were conducted at a constant altitude, but note that the centrifugal force goes by the square of the angular velocity. Also, Mach numbers depend on atmospheric temperature and therefore indirectly on altitude. Strictly speaking, a Mach number for the ISS is meaningless, and even if we are being lax, the figure of Mach 22 is not directly comparable to the X-15 record Mach number. The comparison must be made between speeds, which are, approximately, 17,000 mph for the ISS and 4,500 mph for the X-15 record. If we disregard the relatively small difference in the radius (from the center of the earth), then my BOE calculation suggests that the reduction in weight is about 7%. Note, however, that while the motor is running, its acceleration is the biggest contribution to weight (reaching up to 4g as the fuel weight goes down.) Once it is cut off, then our assumption of level (i.e. non-ballistic) flight assumes that the airplane is capable of generating lift equal to its weight, which implies sufficient atmospheric density for the small wings to generate that lift. This, in turn, seems to imply enough density to produce a non-trivial deceleration due to drag, which also contributes to weight. Therefore, I doubt that there is any phase of the flight where the centrifugal reduction of weight is the dominant effect. [1] https://en.wikipedia.org/wiki/Fictitious_force https://en.wikipedia.org/wiki/Fictitious_force
Svoka 7y ago
It really depends on how you turn. Like, if you would go high enough to have close to none air pressure, and stop accelerating, you would feel about 0% of your body weight, being in free fall, same as ISS. You don't have to go orbital speed to be in free fall. You have to have that speed only not to fall while free falling :)

jacquesm 7y ago

That depends on your orientation.

deleted 7y ago
[deleted]

pdonis 7y ago

> Does that mean that when you're flying an X15 you lose 6.2/22 = 28% of your bodyweight to the Coriolis force? I think you mean centrifugal force. Unfortunately, no, you don't feel quite that much less weight, because centrifugal force goes like the square of the velocity, so the ratio is the square of 6.2/22, or about 0.08, i.e., only 8%.

bzbarsky 7y ago
There is a Coriolis force here too; it's just much smaller than the centrifugal force. Consider a plane flying at ground level along the equator at groundspeed velocity v in the direction of the Earth's rotation. In the Earth's (rotating) reference frame there is a centrifugal acceleration based on the Earth's angular velocity and a Coriolis acceleration given by -2 W x v, where W is the Earth's angular velocity and v is the plane's velocity. This ends up pointing straight up in this configuration. So in the plane's reference frame there is a straight-up centrifugal acceleration, given by (W + v/R)^2 * R = W^2 * R + 2W x v + v^2/R which is balanced by the aerodynamic and gravitational forces on the plane (which in this frame is non-accelerating), while in the Earth's reference frame there is a straight-up centrifugal acceleration given by W^2 * R and a straight-up Coriolis acceleration of 2W x v, and the plane is accelerating radially down at v^2/R due to gravity and lift combined. Now to put numbers to this, the speed record for an X15 is about 2000 m/s, R is 6.4 * 10^6 m, and W is 1 revolution per 24 hours, or 1/86400 s^1. So the three force components look like this, if I am doing my numbers right: W^2 * R is .00086 m/s^2 2W x v is 0.04 m/s^2 v^2/R is 0.63 m/s^2. So the v^2/R "centrifugal force due to the plane's motion" term is by far the biggest contributor here.
- pdonis 7y ago
  > There is a Coriolis force here too; it's just much smaller than the centrifugal force. Yes, agreed.

mannykannot 7y ago

I think that, of the so-called inertial or ficticious forces, you are thinking of the centrifugal rather than the coriolis force [1]. On any of the ballistic / into space flights, you would experience weightlessness between the engine cutoff and the point where atmospheric forces became noticable -- this is what the space tourism companies are aiming for. I do not know if any flights were conducted at a constant altitude, but note that the centrifugal force goes by the square of the angular velocity. Also, Mach numbers depend on atmospheric temperature and therefore indirectly on altitude. Strictly speaking, a Mach number for the ISS is meaningless, and even if we are being lax, the figure of Mach 22 is not directly comparable to the X-15 record Mach number. The comparison must be made between speeds, which are, approximately, 17,000 mph for the ISS and 4,500 mph for the X-15 record. If we disregard the relatively small difference in the radius (from the center of the earth), then my BOE calculation suggests that the reduction in weight is about 7%. Note, however, that while the motor is running, its acceleration is the biggest contribution to weight (reaching up to 4g as the fuel weight goes down.) Once it is cut off, then our assumption of level (i.e. non-ballistic) flight assumes that the airplane is capable of generating lift equal to its weight, which implies sufficient atmospheric density for the small wings to generate that lift. This, in turn, seems to imply enough density to produce a non-trivial deceleration due to drag, which also contributes to weight. Therefore, I doubt that there is any phase of the flight where the centrifugal reduction of weight is the dominant effect. [1] https://en.wikipedia.org/wiki/Fictitious_force https://en.wikipedia.org/wiki/Fictitious_force

Svoka 7y ago

It really depends on how you turn. Like, if you would go high enough to have close to none air pressure, and stop accelerating, you would feel about 0% of your body weight, being in free fall, same as ISS. You don't have to go orbital speed to be in free fall. You have to have that speed only not to fall while free falling :)