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Hi! I need editors who are interested to help clean up Skylon--Novus Orator 09:08, 29 September 2010 (UTC)

Re: AGF

Please read WP:AGF: "This guideline does not require that editors continue to assume good faith in the presence of contrary evidence." Viriditas (talk) 04:44, 30 September 2010 (UTC)

Re: My talk page

Please don't troll my talk page. If you don't have something nice to say, then remain silent. Viriditas (talk) 07:14, 30 September 2010 (UTC)

Physics Discussion about Skylon performance

So this aircraft needs a specially strengthened runway ?? It says so, twice. But if the aircraft has a takeoff mass only one tenth of a loaded Boeing 747, then why does it need a special runway ?Eregli bob (talk) 09:08, 21 September 2010 (UTC)

Simple...Momentum is the product of mass times velocity, and the runway has to deal with that momentum. Though this craft is far lighter than a 747, its takeoff speed will be massive (Half the speed of sound), and will subject the runway to a proportionate amount of stress.--Novus Orator 09:08, 26 September 2010 (UTC)

Well the 747 takes off at 290 km/hour. So "half the speed of sound" is double that. So if the skylon is one tenth of the mass, and double the speed, of the 747, then its momentum is only one fifth of that of the 747. So your rationale doesn't make sense at all.Eregli bob (talk) 13:16, 28 September 2010 (UTC)

747 442 tonnes Skylon 275 tonnes. Half speed of sound 615 km//hour. SO Skylon had 1.32 * the momentum, and 2.8 * the energy.--Kitchen Knife (talk) 14:34, 28 September 2010 (UTC)

I'm not convinced this explanation is sound. Momentum, like velocity, is a vector quantity, and its largest component by far is parallel to the runway. So increasing a plane's takeoff or landing speed (i.e. its groundspeed) doesn't seem relevant to me. As I see it, what's relevant re the necessary strength of the runway is only the momentum that's normal (perpendicular) to the runway, i.e. whether a plane makes a "hard landing" or a "soft" one, and how heavy it is. The vertical component of the velocity at which the plane touches down should be about the same as that of a 747. It would correspond to a vertical airspeed of just a few feet per second, one would hope: anything more and you have a crash landing, anything less and you overshoot and don't leave enough runway for braking after you touch down. The aircraft's groundspeed shouldn't matter at all except as it affects takeoff and landing/braking distance. It's been a long time since college physics, though, so perhaps I've neglected some factor that y'all are aware of?  – OhioStandard (talk) 02:36, 2 October 2010 (UTC)

Yes, you are correct, my explanation was not clear. As far as I understand it, velocity and acceleration are vector quantities, and the acceleration vector of the landing aircraft is diagonal to the runway, e.g. it has a x-component and a y-component. The Force of Kinetic Friction is equal to the Normal Force (Weight) and the mu k (frictional constant), which contribute the additional stress that I am referring to. The y-component is smaller than the x-component, but the Normal Force caused by the aircraft/(in this case spaceplane)'s mass is a part of the frictional force equation (which is the friction the runway exerts on the spaceplane's wheels). The combination of the y-component of the acceleration vector (working down) and the Fk (working horizontally against the wheels) together is what puts the stress on the runway.

To solve for the total linear force the runway must exert on the spaceplane (neglecting air resistance) is:

(in x) (1.0) Frictional Force in x-dimension=(mu k)×(Normal Force)

(in y) (1.1) Normal Force=mass × acceleration

If you wanted to find the Momentum of the Spaceplane once on the runway, you would then take the x-component of the resulting Force in the following equation:

(1.3) Momentum=×(change in time)

Which you could then relate to the horizontal velocity in the Momentum equation:

(1.4) Momentum=mass×velocity

Though this is a very interesting topic, I am reminded of the policy WP:NOTAFORUM and therefore I must end this discussion. It's great knowing that there are editors out there who love Physics as much as I do.. --Novus Orator 03:12, 2 October 2010 (UTC)

Reliable citations

Hi Forums and suchlike are nor reliable to support any content, please have a read of WP:RS Thanks. Off2riorob (talk) 15:13, 2 October 2010 (UTC)

ANI

Hello. This message is being sent to inform you that there currently is a discussion at Misplaced Pages:Administrators' noticeboard/Incidents regarding an issue with which you may have been involved. Thank you. Rd232 13:47, 4 October 2010 (UTC)

And again here. Toddst1 (talk) 07:01, 14 October 2010 (UTC)

And again. Thread restored from archive for additional comment.  – OhioStandard (talk) 16:14, 16 October 2010 (UTC)

Links to Heim theory

Please stop adding links to Heim theory to articles about mainstream science topics (as you did at dark matter and quantum gravity). Per discussions at WT:PHYS, and also at Talk:Heim theory with other editors, Heim theory is sufficiently far from mainstream that linking it from these articles violates WP:UNDUE. The only places links are appropriate are articles that are already discussing fringe or highly-speculative physics models, or articles discussing highly-speculative topics about which Heim theory has been mentioned in high-profile third-party press articles (which is why I left it in at faster than light, and removed it from intergalactic travel).

You seem to understand and acknowledge that Heim theory is far-fringe, so please accept that it won't be linked from the vast majority of physics and space related articles on Misplaced Pages any time soon. That's just one of the aspects of the way Misplaced Pages's science articles work (they follow well-established thought rather than pushing the boundaries). --Christopher Thomas (talk) 19:21, 21 October 2010 (UTC)