Salon has an article on Ingenuity.
In 1903, Orville and Wilbur Wright flew a plane for 12 seconds, 120 feet in the air, on what is now known as the first powered-controlled flight on Earth. Now, 118 years later, the first powered-controlled attempt at a flight on another planet is about to take place.
According to NASA, Ingenuity — the four-pound rotorcraft attached to Perseverance — is on its way to its "airfield" on Mars.
The space agency announced that its target for its first takeoff attempt will happen no earlier than April 8, 2021.
Ingenuity was designed as an experiment to see if it is possible to fly on Mars as we do here on Earth. And the process leading up to the takeoff is a very meticulous one. Consider how long it took humans to stick a powered-controlled flight on Earth; given Mars' thin atmosphere and a twenty-minute delay in communication, it is arguably more challenging on Mars.
"As with everything with the helicopter, this type of deployment has never been done before," Farah Alibay, Mars helicopter integration lead for the Perseverance rover, said in a press statement. "Once we start the deployment there is no turning back."
Every move for the next couple of weeks could make or break Ingenuity's success — starting with precisely positioning the rotorcraft in the middle of its 33-by-33-foot square airfield, which is actually a flat field on the Martian surface with no obstructions. From there, the entire deployment process from Perseverance will take about six Martian days, which are called sols. (The Martian sol is thirty-nine minutes longer than an Earth day.)
Good luck, little chopper!
Previously:
NASA Lays Out Plans for its First Flights on Mars
How NASA Designed a Helicopter that Could Fly Autonomously on Mars
NASA is Sending a Helicopter to Mars, but What For?
(Score: 0) by Anonymous Coward on Monday March 29 2021, @07:21AM (14 children)
It isn't moving anything. It is using the correct units for frequency of cycles/sec instead of 1/sec. While no numerical predictions are altered, this has many consequences such as reinterpreting the "uncertainty principle" as telling us the minimum allowed energy change per cycle. This is described in the link.
(Score: 1) by khallow on Monday March 29 2021, @02:37PM (13 children)
In other words, you moved the unit of cycles from h to f. It doesn't change anything. That numerical predictions aren't altered is an indication that nothing has changed.
In the linked paper above, equation 15 is not based on anything ("Again, using the logic from equation (4), the position-momentum relation is written as"). Look at the difference between equation 11 which is a valid expression of the Heisenberg uncertainty principle and equation 15 which is not: \delta x * \delta p >= h. (11)
\delta x * \delta p >= h_{\delta} * \delta t. (15)
Where did that \delta t come from? Why are we to suppose that h_{\delta} is a constant? Another leap of logic is to then assign \delta E = (\delta x * \delta p)/\delta t (in (17)), and claim that there is a minimum energy step as a result.
The Heisenberg uncertainty principle is a special case of a general idea [wikipedia.org] involving the Fourier transform of noncommuting operators (the momentum p and the energy E are Fourier transforms of the respective position x and time t). This Fourier transform is unique (meaning the transform can be reversed and the inverse Fourier transform also happens to be unique). Energy E already has a Fourier transform t. It can't also have a different second Fourier transform of a constant h. Thus, the inequality of (17) doesn't have a basis in the Fourier transform unlike the other inequalities.
Of course, this is due to a traditional quantum model that is unlikely to hold at extremely small scales of x and t. It is possible that there is a minimum energy step (a discretization of energy) and it may well be your adjusted h in size (as a cycle), but this has not been shown.
Any new, better model, discretized or not, will need to have some sort of Fourier-like transform appearing at the macroscopic level. One big caution is that due to relativity, we are unlikely to have a nice grid pattern of parameters like position and momentum, possibly time and energy too. There will probably be strong limits to what we can discretize.
(Score: 0) by Anonymous Coward on Monday March 29 2021, @03:41PM (12 children)
The SI units of h are J*sec, if you don't even know that I wont bother with the rest.
(Score: 1) by khallow on Monday March 29 2021, @04:28PM (11 children)
(Score: 0) by Anonymous Coward on Monday March 29 2021, @05:29PM (10 children)
How did I move the unit of cycles from where they don't exist?
The problem is that those units are missing from *both* h and f.
(Score: 1) by khallow on Tuesday March 30 2021, @03:57AM (9 children)
Which is fine as long as they are either both missing or both present. Multiply and divide by the same unit cancels.
(Score: 0) by Anonymous Coward on Tuesday March 30 2021, @04:34AM (8 children)
No, it is not fine if the cycle unit is missing. It misleads people to interpret the equations incorrectly.
(Score: 1) by khallow on Tuesday March 30 2021, @08:10AM (7 children)
What's missing about a "cycle unit" that automatically cancels out? It's never present in the first place!
(Score: 0) by Anonymous Coward on Tuesday March 30 2021, @11:46AM (1 child)
It doesn't "automatically cancel out".
We choose the units of the constant so that it does. And those units tell us the meaning of the constant.
(Score: 1) by khallow on Tuesday March 30 2021, @03:23PM
It would not be E= hf, otherwise.
(Score: 0) by Anonymous Coward on Tuesday March 30 2021, @01:24PM (4 children)
By dropping cycles from the units of frequency you have changed the meaning of Planck's constant (and probably others as well).
All the math works out the same but now everyone is confused about what the numbers mean so everything seems "spooky" and non-intuitive.
(Score: 1) by khallow on Tuesday March 30 2021, @03:26PM
All the math works out the same but now everyone is confused about what the numbers mean so everything seems "spooky" and non-intuitive.
"All the math works out the same" says it all. This is irrelevant to the model and the math of that model.
(Score: 1) by khallow on Tuesday March 30 2021, @03:36PM (2 children)
No, we haven't!
"All the math works out the same" says it all. This is irrelevant to the model and the math of that model.
I can't believe there's all this noise over a simple two factor equation, much less in a rocket discussion where even the equation is completely irrelevant. Let me guess, this was all about just dropping a link to that paper?
(Score: 0) by Anonymous Coward on Tuesday March 30 2021, @04:09PM (1 child)
The units of a constant tell you what it represents.
(Score: 1) by khallow on Tuesday March 30 2021, @04:54PM
Again, irrelevant since it doesn't matter if "cycles" is or isn't part of those units.