Difference between revisions of "Talk:Tutorial:Rocket Assembly"
Shadowraven (talk | contribs) (Created page with "Will add screenshots shortly.") |
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| − | Will add screenshots shortly. | + | Will add screenshots shortly. Also, feel free to help with formatting, grammar, or adding pertinent info. --[[User:Shadowraven|shadowraven]] 02:50, 2 July 2014 (UTC) |
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| + | == Removed second paragraph == | ||
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| + | I'm sorry Shadowraven, but you were mistaken on this part (though it was actually useful, so you may want to add a corrected version when you update the tutorial, I don't know how to explain it succinctly). | ||
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| + | τ=r×F, the torque acting on the rocket is a product of both ''radius'' and ''force''. Since the force is generally symmetrical (gravity and vessel acceleration) two payloads of identical weight will still cause a rocket to turn if the payloads' centres of mass are not exactly at the same distance from the rocket's centre of mass. | ||
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| + | However, your point about balancing is very valid when the difference in torque is not so great that it cannot be easily counteracted by the SAS torque. (Or in case of atmospheric boosters by control surface, but why someone would place two different radial boosters is beyond me.) | ||
| + | --[[User:Ghostbird|Ghostbird]] ([[User talk:Ghostbird|talk]]) 17:02, 8 September 2014 (CDT) (heh, I just noticed that our nicknames complement each other) | ||
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| + | Oh, I see there's a note in one of the screenshot texts, so you did consider torque. It is wrong however, since distance DOES affect centre of mass. Imagine a rocket with a 5 ton ballast on a 1m stick on one side and a rocket with a 5 ton ballast on a 100m stick on the other side (consider the sticks weightless). Even when you just imagine this, it is clear that the centre of mass must have moved towards the ballast that is located on the 100m stick. | ||
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| + | This is because the centre of mass is quite literally the ''weighted average'', the average location of all the mass in the ship. Since the ship is now effectively >101m wide, the average location (centre of mass) on this axis has moved towards the average location of the ballast (100m + 1m + 2×ship radius)/2, in the direction of the ~50m mark on the long stick. Considering that the ship will be much heavier than 5 tons, the average location only moves a proportionate distance towards the far ballast.--[[User:Ghostbird|Ghostbird]] ([[User talk:Ghostbird|talk]]) 17:21, 8 September 2014 (CDT) | ||
Latest revision as of 22:24, 8 September 2014
Will add screenshots shortly. Also, feel free to help with formatting, grammar, or adding pertinent info. --shadowraven 02:50, 2 July 2014 (UTC)
Removed second paragraph
I'm sorry Shadowraven, but you were mistaken on this part (though it was actually useful, so you may want to add a corrected version when you update the tutorial, I don't know how to explain it succinctly).
τ=r×F, the torque acting on the rocket is a product of both radius and force. Since the force is generally symmetrical (gravity and vessel acceleration) two payloads of identical weight will still cause a rocket to turn if the payloads' centres of mass are not exactly at the same distance from the rocket's centre of mass.
However, your point about balancing is very valid when the difference in torque is not so great that it cannot be easily counteracted by the SAS torque. (Or in case of atmospheric boosters by control surface, but why someone would place two different radial boosters is beyond me.) --Ghostbird (talk) 17:02, 8 September 2014 (CDT) (heh, I just noticed that our nicknames complement each other)
Oh, I see there's a note in one of the screenshot texts, so you did consider torque. It is wrong however, since distance DOES affect centre of mass. Imagine a rocket with a 5 ton ballast on a 1m stick on one side and a rocket with a 5 ton ballast on a 100m stick on the other side (consider the sticks weightless). Even when you just imagine this, it is clear that the centre of mass must have moved towards the ballast that is located on the 100m stick.
This is because the centre of mass is quite literally the weighted average, the average location of all the mass in the ship. Since the ship is now effectively >101m wide, the average location (centre of mass) on this axis has moved towards the average location of the ballast (100m + 1m + 2×ship radius)/2, in the direction of the ~50m mark on the long stick. Considering that the ship will be much heavier than 5 tons, the average location only moves a proportionate distance towards the far ballast.--Ghostbird (talk) 17:21, 8 September 2014 (CDT)
