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Forum » SpaceEngine » Feedback and Suggestions » General suggestions (Post your suggestions here.)
General suggestions
JackDoleDate: Monday, 08.02.2016, 13:23 | Message # 691
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An urgent request: biggrin

Will there be a function in the foreseeable future, with which one can update a script in SpaceEngine during operation?

Through a console script? As an example, if I have a number of scripts that show a different level of development of a star?
Then they could be activated by a console script sequentially.

A shortcut would be enough already.
I could then replace the .sc scripts with a batch script and activate with AutoHotkey!





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PlutonianEmpireDate: Tuesday, 09.02.2016, 06:50 | Message # 692
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Would it be possible to give ships an internal gyroscope, in the sense that if the ship is in orbit around the planet, and the user sets the prograde or horizon button for example, the ship will automatically maintain the requested relative orientation without having to fire thrusters every two seconds?




Specs: Dell Inspiron 5547 (Laptop); 8 gigabytes of RAM; Processor: Intel® Core™ i5-4210U CPU @ 1.70GHz (4 CPUs), ~2.4GHz; Operating System: Windows 7 Home Premium 64-bit; Graphics: Intel® HD Graphics 4400 (That's all there is :( )

Edited by PlutonianEmpire - Tuesday, 09.02.2016, 06:52
 
SpaceEngineerDate: Tuesday, 09.02.2016, 07:38 | Message # 693
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Quote JackDole ()
Will there be a function in the foreseeable future, with which one can update a script in SpaceEngine during operation?

This is impossible in current engine architecture. It require fully dynamic internal database structure.

Quote PlutonianEmpire ()
Would it be possible to give ships an internal gyroscope, in the sense that if the ship is in orbit around the planet, and the user sets the prograde or horizon button for example, the ship will automatically maintain the requested relative orientation without having to fire thrusters every two seconds?

This will be in the future, but now it's more easy to repeat prograde comment each few seconds.





 
JackDoleDate: Tuesday, 09.02.2016, 09:13 | Message # 694
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Quote SpaceEngineer ()
This is impossible in current engine architecture. It require fully dynamic internal database structure.

That I was afraid of. sad

The thought came to me afterwards that SpaceEngine then would try every time to re-read the entire database, and that the import of a single script probably would be very complicated.



Would it be possible to create a function that makes an object 'invisible'?
Something like: 'NoVisible true'?

That this switch then for a specific object could be set to 'NoVisible false' with a console command. And vice versa.

Would something like this be very complicated?





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Edited by JackDole - Tuesday, 09.02.2016, 09:41
 
CanleskisDate: Thursday, 18.02.2016, 22:01 | Message # 695
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I have a suggestion and this is something I would really like to see. I don't know why, but I have some troubles to determinate the distance between two things in Space Engine, I assume it is not an unnatural thing because of the vastness of the universe, but when for example I am on a mountain on a planet, I cannot determinate the distance with this mountain and another one, it confused me so much that I can't really appreciate the landscape sad The thing I would really like to see is similar to what Outerra does, the game (or the engine actually) shows the distance between you and another place according to the placement of the crosshair. Here's some examples:









I don't really know if it is possible for the moment to make somethink like this in Space Engine, but I think that's a good idea and could help me and maybe everyone if I am not the only person having this weird sensation wacko
One last question by the way, why does the Moon looks much bigger in Outerra than in Space Engine whereas they are at the same distance? (of course with the same FOV, I tested it)

Attachments: 2560421.jpg(246Kb) · 3459100.jpg(443Kb) · 7396125.jpg(322Kb) · 8309686.png(258Kb)


Edited by Canleskis - Thursday, 18.02.2016, 22:05
 
quarior14Date: Friday, 19.02.2016, 14:22 | Message # 696
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With the new update of Space Engine, you can choose the unit of the atmosphere so I suggest if you can also with gravity or there will be either g or N·m²·kg−2 or m^3·kg−1·s^−2.
Reminder : 1 g = 9.81 N·m²·kg^−2

Also, is that it is possible to display the "surface" gravity event horizon or surface of the tidal force by distance of black holes ?
Screenshot (sorry, it is French) :


Also, what is a way for to docking automatically because it already has auto-pilot to reach a destination.

Attachments: 2923514.png(91Kb)





Quarior

Edited by quarior14 - Friday, 19.02.2016, 20:19
 
SpaceEngineerDate: Saturday, 20.02.2016, 08:12 | Message # 697
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Quote Canleskis ()
shows the distance between you and another place according to the placement of the crosshair. Here's some examples:

This is not easy to implement in SE.

Quote quarior14 ()
Also, is that it is possible to display the "surface" gravity event horizon or surface of the tidal force by distance of black holes ?

I don't understand, please make correct sentence.

Quote quarior14 ()
Also, what is a way for to docking automatically because it already has auto-pilot to reach a destination.

Automatic docking is not implemented yet. Using cheats (EnableCheats true) it is possible to dock instantly.





 
WatsisnameDate: Saturday, 20.02.2016, 11:20 | Message # 698
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Quote quarior14 ()
Reminder : 1 g = 9.81 N·m²·kg^−2


You are confusing the units of "g" (acceleration due to gravity at Earth's surface, which averages 9.81 m/s2), with "G" (the gravitational constant, 6.67*10-11 N*m2/kg2).

I don't understand your suggestion or questions either. If you're asking if it's possible to show the surface gravity of a black hole (at its horizon), then this is a meaningless quantity. In some respects it is infinite.

If you're asking if it's possible to show the tidal forces from a black hole, this is already shown, in units of g per meter. You can consider a tidal force stronger than (very roughly) 10g/m to be lethal.





 
CanleskisDate: Saturday, 20.02.2016, 19:18 | Message # 699
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Quote SpaceEngineer ()
This is not easy to implement in SE.


This is what I expected, thanks for the answer but I hope it will be implemented one day smile
 
quarior14Date: Wednesday, 24.02.2016, 15:49 | Message # 700
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Quote Watsisname ()
Quote quarior14 ()
1 g = 9.81 N·m²·kg^−2

You are confusing the units of "g" (acceleration due to gravity at Earth's surface, which averages 9.81 m/s2), with "G" (the gravitational constant, 6.67*10-11 N*m2/kg2).I don't understand your suggestion or questions either. If you're asking if it's possible to show the surface gravity of a black hole (at its horizon), then this is a meaningless quantity. In some respects it is infinite.If you're asking if it's possible to show the tidal forces from a black hole, this is already shown, in units of g per meter. You can consider a tidal force stronger than (very roughly) 10g/m to be lethal.

Yes, it is in fact N.kg^-1 but what is the relationship between the strength of the tidal force (g/m) ?
Quote SpaceEngineer ()
Quote quarior14 ()
Also, is that it is possible to display the "surface" gravity event horizon or surface of the tidal force by distance of black holes ?

I don't understand, please make correct sentence.

I use with the following formula :
g = (G*m)/d²
g in N.kg^-1 (gravity)
G = 6.67*10^(-11) N.m²/kg² (gravitational constant)
m in kg (mass)
d in m (distance)
Exemple for Sagittarius A* (Mass : 4.31*10^6 * 1.977*10^(30) = 8.5769*10^(36) kg, Radius (event horizon) : 0.085 AU = 1.27*10^(10) m) :
g = (6.67*10^(-11)*8.5769*10^(36))/(1.27*10^(10))² = 1.29*10^47 N.kg^-1 = 1.31*10^46 g
I hope you understand, I do not know how to describe further, d it is the distance between camera and the center of black hole.





Quarior

Edited by quarior14 - Wednesday, 24.02.2016, 15:50
 
Kubacki99Date: Friday, 26.02.2016, 22:06 | Message # 701
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Is it possible to add some function, which shows selected star on the Hertzsprung–Russell diagram?
 
WatsisnameDate: Saturday, 27.02.2016, 01:28 | Message # 702
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Quote quarior14
Yes, it is in fact N.kg^-1 but what is the relationship between the strength of the tidal force (g/m) ?


Correct, or just m/s2. Acceleration due to gravity is independent of the mass, so why use force per mass?

Tidal force is the change in force of gravity with respect to distance. To calculate it with Newton's Laws:

g(d)-g(d+Δd)=GM(1/d2-1/(d+Δd)2)

Quote quarior14 ()
Exemple for Sagittarius A* (Mass : 4.31*10^6 * 1.977*10^(30) = 8.5769*10^(36) kg, Radius (event horizon) : 0.085 AU = 1.27*10^(10) m) :
g = (6.67*10^(-11)*8.5769*10^(36))/(1.27*10^(10))² = 1.29*10^47 N.kg^-1 = 1.31*10^46 g


Uhh, check your math. You should be getting something like 3.5*106m/s2, or about 360,000 times Earth's surface gravity. Also, since event horizon radius (of a Schwarzschild black hole) is 2GM/c2, you can calculate it as
g(event horizon) = c4/(4GM)

However, this calculation is wrong. It's using Newton's laws in a *very* general relativistic regime. The result suggests you could hover at the event horizon if you accelerate upward with finite acceleration. This is wrong. The acceleration needed would actually be infinite. Only photons or other massless particles moving at the speed of light can remain on the horizon.

So... I wouldn't use this calculation. There are other interpretations of "surface gravity" of a black hole that mimic it (particularly in how it depends on the mass of the hole), but it's really not the same thing.

Similarly, the calculation of tidal force given above isn't completely accurate near a black hole. SE calculates it general relativistically (at least I'm pretty sure) from the space-time curvature.

Edit: Sorry, quoted wrong person.





 
quarior14Date: Saturday, 27.02.2016, 10:04 | Message # 703
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Quote Watsisname ()
Correct, or just m/s². Acceleration due to gravity is independent of the mass, so why use force per mass?

It is the unity international.
Quote Watsisname ()
Uhh, check your math. You should be getting something like 3.5*10^6 m/s²

Ah yes, i forget the square, it is g = (6.67*10^(-11)*8.5769*10^(36))/(1.27*10^(10))² = 3546898.32 m/s² (or N/kg) = 361559.4618 g
Quote Watsisname ()
g(event horizon) = c^4/(4GM)

M = 8.5769*10^(36) kg
c = 299 792 458 m/s
G = 6.67*10^(-11) N.m²/kg²
g(event horizon) = (299792458)^4/(4*6.67*10^(-11)*(8.5769*10^(36)) = 3529934.443 m/s² = 359830.2185 g
Damn, it looks like my previous result.
In fact what is strange is that when the distance is 0 as gravity takes infinite + from the relationship (watch over limit in mathematics) to any body even if in reality, I'm not on it's realistic except for black hole I think.
Quote Watsisname ()
However, this calculation is wrong. It's using Newton's laws in a *very* general relativistic regime. The result suggests you could hover at the event horizon if you accelerate upward with finite acceleration. This is wrong. The acceleration needed would actually be infinite. Only photons or other massless particles moving at the speed of light can remain on the horizon. So... I wouldn't use this calculation. There are other interpretations of "surface gravity" of a black hole that mimic it (particularly in how it depends on the mass of the hole), but it's really not the same thing.Similarly, the calculation of tidal force given above isn't completely accurate near a black hole. SE calculates it general relativistically (at least I'm pretty sure) from the space-time curvature.

I think there is beyond my knowledge so I do not know what to say.

PS : How do you put the other powers that ² because even by copy paste, put it as a figure "normal" (example 4) ?





Quarior

Edited by quarior14 - Saturday, 27.02.2016, 10:06
 
HarbingerDawnDate: Saturday, 27.02.2016, 17:59 | Message # 704
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Quote quarior14 ()
How do you put the other powers that ² because even by copy paste, put it as a figure "normal" (example 4) ?

Code
10[sup]4[/sup] m/s[sup]2[/sup]

produces

104 m/s2





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Edited by HarbingerDawn - Saturday, 27.02.2016, 18:00
 
WatsisnameDate: Saturday, 27.02.2016, 22:04 | Message # 705
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Quote quarior14 ()
Damn, it looks like my previous result.


Yeah, it should. smile It's the same formula, just expressed differently. It replaced distance with the event horizon radius.

Quote quarior14 ()
In fact what is strange is that when the distance is 0 as gravity takes infinite

Of course. The limit of 1/r2 as r goes to zero is infinity. If a black hole singularity is infinitely small, then it is infinitely attractive at that point.

Quote quarior14 ()

I think there is beyond my knowledge so I do not know what to say.


Basically all I am saying is that the formula for the force of gravity F=GMm/r2 (Newton's Law of Gravity) is not accurate near a black hole. We really need to use general relativity.





 
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