Proposal Discussion Flight Assist Off

[Globusdiablo]

I messaged this directly to GlobusDiablo, but I thought I'd post it here for Mike et al to read as it may clear up the confusion:

I believe what is being described is an illusion related to the enforced top speed in ED.

Even in Flight Assist Off mode you are limited by a top speed.
When you push and hold the stick half way forward this slowly accelerates you until you reach your top speed (say 160).
When you turn while maintaining the stick half way forward then it does indeed start altering your ship's trajectory. However, because of the top speed, the readout remains reading 160.

Think of your ship as being in the middle of a 'speed bubble', you are remaining on the edge of that bubble, so your momentum reading remains at 160, even though the direction you are travelling is slowly adjusting. Even with Flight Assist Off, Flight Limiters remain on, and the engines provide a counter thrust preventing you from straying over that 160. So if you turned straight up 90 degrees then you would eventually transition from [160,0,0] to [80,80,0] to [0,160,0], slowly losing your original forward momentum in favour of your new direction. The speed reading would remain at 160 throughout this.

But, if you turn quickly to face the opposite direction, then you'll see the speed reading change, dropping from 160 down and then back up to 160 as it crosses through the centre of the 'speed bubble' (rather than scraping round the edge of it and maintaining the 160 reading like it does when you turn slowly or at an angle less than 180 degrees).

This is all correct behaviour when taking the speed limit cap into account.

Thanks for getting back to me Isinona

You may be right. But I don't think so. In the described scenarios, when I change the heading of the ship, nothing happens no matter how long I wait (keep floating sideways) unless I do a 180. But you could be right. Because of the speed, the turn radius could be so huge, that it doesn't register in the cockpit window, thus never flying truly forwards... But somehow it doesn't feel right.
But Premium Beta is tomorrow, so I have fingers and toes crossed for a different experience in the updated build...

 

[Dejay]

Interesting. You probably also should use a mix of foward / lateral thrusters to quickly change your trajectory.

 

[Globusdiablo]

Interesting. You probably also should use a mix of foward / lateral thrusters to quickly change your trajectory.

Hey Dejay

Yeah, I know. But I think the problem is that FAOFF just isn't supported well enough for the 360 controller. Being able to preset a speed (50%) as with a throttle stick is absolutely necessary, otherwise FAOFF with the Xbox controller is, in my opinion, unplayable.

But lets see what premium beta brings... <are we there yet?..>

 

[Viajero]

So if you turned straight up 90 degrees then you would eventually transition from [160,0,0] to [80,80,0] to [0,160,0], slowly losing your original forward momentum in favour of your new direction. The speed reading would remain at 160 throughout this.

Great post Isinona (also I´m an avid watcher and fan of your FA OFF videos!)

BTW, if my rusty maths and physics dont fail me that middle stage [80,80,0] may be technically incorrect as it would represent an absolute velocity value of just:

SQR (80^2+80^2) = 113, and not 160.

That is assuming the flight model tries to maintain that max 160 reading as absolute velocity irrespective of the orientation of the ship?

Actual transition at equal x and y velocites would then rather be

SQR((160^2)/2) = 113 (funnily enough)

So [160,0,0] to [113,113,0] to [0,160,0]

 

[Isinona]

Great post Isinona (also I´m an avid watcher and fan of your FA OFF videos!)

BTW, if my rusty maths and physics dont fail me that middle stage [80,80,0] may be technically incorrect as it would represent an absolute velocity value of just:

SQR (80^2+80^2) = 113, and not 160.

That is assuming the flight model tries to maintain that max 160 reading as absolute velocity irrespective of the orientation of the ship?

Actual transition at equal x and y velocites would then rather be

SQR((160^2)/2) = 113 (funnily enough)

So [160,0,0] to [113,113,0] to [0,160,0]

Ha, yeah you're totally right, Pythagoras would be disappointed! Some lazy wrong maths by me there.