Difference between revisions of "Talk:Atmosphere"
From Kerbal Space Program Wiki
(Explanation of the purpose of weight in real-world drag.) |
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== Drag calculation notes == | == Drag calculation notes == | ||
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The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it. | The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it. | ||
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation. | Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation. | ||
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| − | Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? --[[User:T0w0i7ne|T0w0i7ne]] | + | Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? <nowiki>--~~~~</nowiki>. — [[User:T0w0i7ne|T0w0i7ne]] <small>[[[User talk:T0w0i7ne|talk]]]</small> 13:00 (ish), 15 July 2013 (MDT) |
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:First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>[[[User talk:XZise|talk]]]</small> 16:26, 15 July 2013 (CDT) | :First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the ''m'' term is not present in the [[w:drag_equation|real-word drag equation]]. In the game, this causes acceleration due to drag (''a = FD / m'') to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with <nowiki>--~~~~</nowiki>. — [[User:XZise|xZise]] <small>[[[User talk:XZise|talk]]]</small> 16:26, 15 July 2013 (CDT) | ||
| − | ::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --[[User:T0w0i7ne|T0w0i7ne]] | + | |
| − | ::: | + | ::I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. <nowiki>--~~~~</nowiki>. — [[User:T0w0i7ne|T0w0i7ne]] <small>[[[User talk:T0w0i7ne|talk]]]</small> 21:37, 15 July 2013 (MDT) |
| + | :::Actually, the force of drag IS independent of the mass of the craft, however, the acceleration is not. This is because mass of the object is used in conversion between force applied, in newtons, and actual acceleration, in m/s².--[[User:Ruedii|Ruedii]] ([[User talk:Ruedii|talk]]) 17:49, 16 July 2013 (CDT) | ||
Revision as of 22:49, 16 July 2013
Drag calculation notes
The calculation for the drag force as listed on this wiki page is incorrect. The units for the equation provided would be (Kg^2*m)/(s^2) when the unit of force (N) is Kg*m/(s^s). The equation should not include the mass of the craft as the mass of the craft has nothing to do with the amount of drag exerted on it.
Remember that F=ma is a very simplified version of Newton's second law. The real way to read the equation is the sum of all instantaneous forces on a body is equal to the instantaneous mass of the body multiplied by the instantaneous acceleration of the body. It is actually more helpful to think of this drag force calculation in the context of Newton's third law.The craft is accelerating the air particles it contacts (or the pressure zone ahead of it contacts) and the equal but opposite force is applied to the craft. The mass that matters when determining aerodynamic forces is the mass of the disturbed air and this is why the density of the disturbed fluid is part of the calculation.
Edit:
Before writing this I did not see the mention to this fact in the wiki page, but I am still not sure why a force would be modeled with completely wrong units. It is not to make the amount of drag be independent of the craft's mass because, as I explained above the drag equation does not consider the craft's mass. Are the internal modeling calculations all done in mass specific forces/accelerations? If so, shouldn't the true drag force then be divided by the craft's mass to yield this value? --~~~~. — T0w0i7ne [talk] 13:00 (ish), 15 July 2013 (MDT)
- First of all, this biiiig paragraph isn't very readable. I actually only read some of it and I'm asking you: Do you complain, that the formula here is not real world compatible? Then I reference to: “Note that the m term is not present in the real-word drag equation. In the game, this causes acceleration due to drag (a = FD / m) to be unaffected by a ship's mass. (It also causes the units of the drag equation to have an extra "kg" term.)”. By the way, please sign your posts with --~~~~. — xZise [talk] 16:26, 15 July 2013 (CDT)
- I misread that section once I saw it. I now understand the desire was to make the acceleration of the craft due to the force of drag be independent of the mass of the craft. What is the point in such an approach when the reality is precisely the opposite? What does this gain in the simulation? I don't see that this would be any simpler to code or that it would use drastically less CPU cycles. Please excuse my ignorance of wiki etiquette, I am now in the process of learning how to make nicer posts (after an attempt to clean up the above atrocity). I was originally under the impression that this discussion page was more like forum postings. --~~~~. — T0w0i7ne [talk] 21:37, 15 July 2013 (MDT)
