5 ms·

If you're zeroing in the position, you don't have to worry about the motion. The number of degrees you have to take car of is 6: X, Y, Z, yaw, pitch, and roll

If you're zeroing in the position, you don't have to worry about the motion. The number of degrees you have to take car of is 6: X, Y, Z, yaw, pitch, and roll.

JshWright 11y ago
Yes, but the only means you have of controlling those also impacts the motion.
- Animats 11y ago
  Yes. Welcome to underactuated control, where you have more output goals than control parameters. That's why this is rocket science. Here's a way to start thinking about this. Consider the 1D case, stopping a car with the goal of being at zero speed at a specified point. 2 goals, one controlled input. This is an under actuated problem. Setting a constant deceleration is not enough to do this. However, if you have a deceleration start time and a deceleration value after that point, you now have two variables, and can solve for zero speed at the desired point. This is essentially what Space-X is doing in the vertical direction. They have limited ability to throttle the main engines (off, or 70% to 100%), and I suspect that in the final landing phase, they use that range to keep the vertical component constant while doing any horizontal positioning. The main engines gimbal; they don't have to change attitude for minor horizontal adjustments. So the simple approach is to get the attitude stabilized on approach, and the landing vertical. Their successful landings look like that. Their unsuccessful ones show non-vertical attitudes as the control system tries to make big horizontal adjustments.
  - JshWright 11y ago
    This is also why New Shepard (with a minimum TWR < 1) has a much easier problem to solve (on the scale of 'solving problems related to rockets).
drjesusphd 11y ago
> If you're zeroing in the position, you don't have to worry about the motion. Sure you do. Landing too fast in any of the three directions won't do.
dzdt 11y ago
If you don't take care of the motion, the next thing you do immediately after landing on the barge is slide off the barge!
- mikeash 11y ago
  Or make a big hole in it, as happened last time.
sehugg 11y ago
Heck, I'll just be pedantic and say you have 7 DOF since you really should use a quaternion to avoid the gimbal-lock singularity when using Euler angles :P
- visarga 11y ago
  To a layman the two are nearly identical, however the Von Hiesen is an older model of Neutrino Compressors and thus shares many aesthetic components as the contemporary hypercarbolators (I tend to use the American spelling for that word) due to the placement of the backfeed tube.
  - chronolitus 11y ago
    gimbal and quaternions are two different ways of describing an object's rotation in 3d. gimbal is very intuitive, you just write down three angles: pitch, roll, and yaw. to imagine the objects rotation when I give you the three angles, you just apply them one after the other. however, there is an issue. When the object's rotation is large enough that say, roll becomes 90, suddenly pitch and yaw correspond to the same 'thing '. which means one axis of rotation can no longer be realized. in practice this can be seen with physical gimbals (three concentric rings, one per axis) or in numerical gimbal representation, where values might end up being unworkable. quaternions eliminate this problem by using four values instead of three (quater, quattro, quad bike - the prefix stands for 4) 3 values are the components of a 3d vector, and the 4th is how much rotation to apply around it. The above post argue that this 4th value seems like an extra degree of freedom, it isnt. it simply is a matter of representation if coded correctly.
- julbaxter 11y ago
  Using quaternion does not add any DOF.
  - sehugg 11y ago
    You're right, assuming your quaternions are coded correctly...
static_noise 11y ago
Yes, 6 positions, but you also have to take of their first derivative (velocity) which makes it 12.

If you're zeroing in the position, you don't have to worry about the motion. The number of degrees you have to take car of is 6: X, Y, Z, yaw, pitch, and roll.

JshWright 11y ago

Yes, but the only means you have of controlling those also impacts the motion.

Animats 11y ago
Yes. Welcome to underactuated control, where you have more output goals than control parameters. That's why this is rocket science. Here's a way to start thinking about this. Consider the 1D case, stopping a car with the goal of being at zero speed at a specified point. 2 goals, one controlled input. This is an under actuated problem. Setting a constant deceleration is not enough to do this. However, if you have a deceleration start time and a deceleration value after that point, you now have two variables, and can solve for zero speed at the desired point. This is essentially what Space-X is doing in the vertical direction. They have limited ability to throttle the main engines (off, or 70% to 100%), and I suspect that in the final landing phase, they use that range to keep the vertical component constant while doing any horizontal positioning. The main engines gimbal; they don't have to change attitude for minor horizontal adjustments. So the simple approach is to get the attitude stabilized on approach, and the landing vertical. Their successful landings look like that. Their unsuccessful ones show non-vertical attitudes as the control system tries to make big horizontal adjustments.
- JshWright 11y ago
  This is also why New Shepard (with a minimum TWR < 1) has a much easier problem to solve (on the scale of 'solving problems related to rockets).

drjesusphd 11y ago

> If you're zeroing in the position, you don't have to worry about the motion. Sure you do. Landing too fast in any of the three directions won't do.

dzdt 11y ago

If you don't take care of the motion, the next thing you do immediately after landing on the barge is slide off the barge!

mikeash 11y ago
Or make a big hole in it, as happened last time.

sehugg 11y ago

Heck, I'll just be pedantic and say you have 7 DOF since you really should use a quaternion to avoid the gimbal-lock singularity when using Euler angles :P

visarga 11y ago
To a layman the two are nearly identical, however the Von Hiesen is an older model of Neutrino Compressors and thus shares many aesthetic components as the contemporary hypercarbolators (I tend to use the American spelling for that word) due to the placement of the backfeed tube.
- chronolitus 11y ago
  gimbal and quaternions are two different ways of describing an object's rotation in 3d. gimbal is very intuitive, you just write down three angles: pitch, roll, and yaw. to imagine the objects rotation when I give you the three angles, you just apply them one after the other. however, there is an issue. When the object's rotation is large enough that say, roll becomes 90, suddenly pitch and yaw correspond to the same 'thing '. which means one axis of rotation can no longer be realized. in practice this can be seen with physical gimbals (three concentric rings, one per axis) or in numerical gimbal representation, where values might end up being unworkable. quaternions eliminate this problem by using four values instead of three (quater, quattro, quad bike - the prefix stands for 4) 3 values are the components of a 3d vector, and the 4th is how much rotation to apply around it. The above post argue that this 4th value seems like an extra degree of freedom, it isnt. it simply is a matter of representation if coded correctly.
julbaxter 11y ago
Using quaternion does not add any DOF.
- sehugg 11y ago
  You're right, assuming your quaternions are coded correctly...

static_noise 11y ago

Yes, 6 positions, but you also have to take of their first derivative (velocity) which makes it 12.