[chabig]

The issue isn't whether the forces are added together. It's a human factors issue about making sure the pilots know what the other is doing.

[chabig]

Originally Posted by imitchellg5 

Not necessarily. Just more pilot in command time.

Not true. It's a seniority system. The pilot who got hired first is more senior. Period. It's not based on PIC time at all.

[driven]

Originally Posted by chabig 

Once again, it doesn't average. It adds the inputs together. And though we agree that you can make up situations where it might not make sense, I leave it to you to propose a better solution. I can make up a situation in a mechanical yoked airplane where everyone dies. Here you go:

You are approaching a light aircraft which you need to avoid.
One pilot choses to descend and pushes forward on the yoke
One pilot choses to climb and pulls back on the yoke with equal force

The yokes don't move and everyone dies.

Do you see? It's exactly the same! The forces were algebraically added together, exactly as they are in the fly-by-wire system.

Well, not exactly the same.

In your situation at least one of the pilots would say to the other "dude, let's get together here." In the fly-by-wire situation, it could go un-noticed. (Indeed, it did)

[chabig]

That's what I said in the next post.

[imitchellg5]

Originally Posted by chabig 

Not true. It's a seniority system. The pilot who got hired first is more senior. Period. It's not based on PIC time at all.

For American and some European airlines, yes. But other airline, such as Emirates, are based strictly upon PIC.

[chabig]

That seems odd to me. So you're saying two pilots are scheduled to fly together, they show up for the flight, compare PIC time and the high time guy sits on the left? Do you only count company time or do you compare total time in all flight operations, military, civilian, etc? If what you say it correct, why does Emirates have Captains? Their website discusses Captain versus FO pay, for example, so there is a distinction. They have 2909 pilots today. If what you say is correct and I'm an Emirates pilot with 10 years of service, and they hire a new guy who has slightly more flight time than me, I'd be pissed if he gets the left seat instead of me. I doubt the accuracy of your assertion.

[imitchellg5]

Originally Posted by chabig 

That seems odd to me. So you're saying two pilots are scheduled to fly together, they show up for the flight, compare PIC time and the high time guy sits on the left? Do you only count company time or do you compare total time in all flight operations, military, civilian, etc?

Uh. No. The one with the rank of captain sits of the left, the one who's a copilot sits on the right.

For example, one of my dad's good friends just got hired by Cathay even though he's a UA 320 copilot for the left seat in the A333 because he had higher PIC time compared to the rest of his hiring class.

Emirates is in a binge right now of hiring US pilots with high PIC time to start as pilots in the 77W regardless of whether or not they are pilot or copilot at their current jobs.

[chabig]

Oh right. I know what you're talking about. Some of those rapidly growing airlines don't have enough experienced pilots to flow them into the left seat, so they hire more experienced pilots to start as Captains. So yes, it's not fully seniority based. I forgot about that. It's a result of the airline's rate of growth.

[imitchellg5]

Well, that, and having a good amount of high-time pilots who want to get out of the messed up US airline industry.

[11011001]

Originally Posted by chabig 

Once again, it doesn't average. It adds the inputs together. And though we agree that you can make up situations where it might not make sense, I leave it to you to propose a better solution. I can make up a situation in a mechanical yoked airplane where everyone dies. Here you go:

You are approaching a light aircraft which you need to avoid.
One pilot choses to descend and pushes forward on the yoke
One pilot choses to climb and pulls back on the yoke with equal force

The yokes don't move and everyone dies.

Do you see? It's exactly the same! The forces were algebraically added together, exactly as they are in the fly-by-wire system.

What about a control system that is mutually exclusive. A button on the stick gives it control, and the other stick becomes mechanically locked in place, giving the pilots a clear and intuitive indication of who is in control. Do any fly-by-wire planes do this?

Why did they ever average the inputs? Was it thought to be intuitive, because of the way a mechanically yoked plane operated?

[chabig]

Originally Posted by 11011001

What about a control system that is mutually exclusive. A button on the stick gives it control, and the other stick becomes mechanically locked in place, giving the pilots a clear and intuitive indication of who is in control. Do any fly-by-wire planes do this?

Not that I know of, but the concept would be worth exploring, it it hasn't already. An extra button to take control would be a big nuisance, though. You should not normally have to push a button to have control.

Originally Posted by 11011001

Why did they ever average the inputs?

They never averaged the inputs.

[11011001]

Originally Posted by chabig 

They never averaged the inputs.

Originally Posted by driven 

- Airbus averages the two conflicting inputs for the actual changes to the control surfaces.

What did they do then? Addition of two vectors (assume these are transposed):
(0,-1) + (0,1) = (0,0)
(0, 1) + (0,1) = (0,2) <- summing doesn't work here

Averaging:
((0, -1) + (0,1))*0.5 = (0,0)
((0, 1) + (0,1))*0.5 = (0,1) <- what we would expect

[chabig]

Driven was mistaken. The inputs are added, as in 1+1=2. Of course, the maximum is limited to 1. To copy your examples:

(0, .1) + (0, .1) = (0, .2)
(0, .1) + (0, .2) = (0, .3)
(0, .2) + (0, .2) = (0, .4)
(0, .2) + (0, .3) = (0, .5)
(0, .3) + (0, .3) = (0, .6)
(0, .3) + (0, .4) = (0, .7)
(0, .3) + (0, .5) = (0, .8)
(0, .3) + (0, .6) = (0, .9)
(0, .3) + (0, .7) = (0, 1.0)
(0, .3) + (0, .8) = (0, 1.0) <- output limited to 1
.
.
(0, .9) + (0, .9) = (0, 1.0)
(0, 1.0) + (0, 1.0) = (0, 1.0)

[11011001]

For input a, and b, the output, o is:
o = (min(a1 + b1,1), min(a2 + b2,1))

gotcha.

edit:
o = (min(max(a1 + b1,-1), 1), min(max(a2 + b2, -1), 1))

[chabig]

I have a feeling that you won't take my word as an expert, so here is a page from the documentation:

[chabig]

Originally Posted by 11011001 

For input a, and b, the output, o is:
o = (min(a1 + b1,1), min(a2 + b2,1))

gotcha.

Yeah. That looks good.

[11011001]

Oh, hmm, we'd need something more to handle negative numbers, but yea, I understand.

[driven]

What if both aviators push the "take control" button? Which stick controls the aircraft?

[chabig]

See the last sentence on the page I linked. Whomever pushes it last has control.