User talk:TASDELEN

Welcome

Hello TASDELEN and welcome to Wikipedia! I am Ukexpat and I would like to thank you for your contributions.

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ukexpat (talk) 20:57, 8 February 2009 (UTC)[reply]

May 2010

Welcome to Wikipedia! I am glad to see you are interested in discussing a topic. However, as a general rule, talk pages such as Ellipse are for discussion related to improving the article, not general discussion about the topic. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. Thank you. The Evil IP address (talk) 18:01, 3 May 2010 (UTC)[reply]

No OR

Please don't add your OR to any more articles Nil Einne (talk) 10:15, 6 May 2010 (UTC)[reply]

Please stop

Please do not add unsourced or original content, as you did to Ellipse. Doing so violates Wikipedia's verifiability policy. If you continue to do so, you may be blocked from editing Wikipedia. - DVdm (talk) 12:25, 10 May 2010 (UTC)[reply]

Sorry,Please delete the add,or transporte it to "discussion page".I could not find how to delete or how to open a discussion page.TASDELEN (talk) 12:31, 10 May 2010 (UTC)[reply]

It does not belong on the discussion page either. Please read the talk page guidelines. DVdm (talk) 12:33, 10 May 2010 (UTC)[reply]

This is the final warning you will receive regarding your disruptive edits.
If you continue to use talk pages such as Talk:Ellipse for inappropriate discussions you may be blocked from editing without further notice.

See User talk:78.191.97.74 - DVdm (talk) 10:11, 11 May 2010 (UTC)[reply]

The article I added was not an editing.It was a new topic page.Now the same article is on "mathematic reference desk".If you wish to delete it,please do it,now.I am not an editor,but I want to be edited.Thanks.TASDELEN (talk) 13:13, 11 May 2010 (UTC)[reply]

Reference desk

I have removed this post by you on the reference desk. Your understanding of mechanics is fundamentally flawed, as we have tried to explain to you. Please do not confuse other people by spouting your nonsense in answer to their questions. --Tango (talk) 00:42, 25 May 2010 (UTC)[reply]

I want to add this: Decide whether you are asking or answering. Asking is acknowledging ignorance. Answering is referring to somebody elses question. As you indicate that you are right and everybody else is wrong, you are neither asking nor answering, but preaching, and that is not what the reference desk is about, and you are loosing respect.

Re you question: Consider the special case where the force of gravitation is zero. Then the planet moves straight pass the sun with constant speed V and minimum distance a, and the area speed rV_p = aV is constant too, but the angular speed V_p = aV/r varies with the distance r. Bo Jacoby (talk) 11:59, 2 June 2010 (UTC).[reply]

I don't care loosing respect in the reasoning of copy-paste minded mathematicians. What I am looking for is if Newton's law F*dt=m*dv is agreeed or refused. If agreed then dVp/dt=0. Then comes the question: why Wiki says d(r*Vp)/dt=0 ? I aknowledge my ignorance about this. Thanks. TASDELEN (talk) 19:04, 2 June 2010 (UTC)[reply]

I think the flaws in your argument have been clearly explained to you by experts on the Mathematics Reference Desk, and I tried to explain to you on the Science Desk that a theory is useful only if it fits the observations. If you insist on your one-person campaign to overturn Newton, Kepler etc, then why not start a blog where you can say what you like without disrupting Wikipedia? This is not the place for original research. Dbfirs 07:49, 9 June 2010 (UTC)[reply]

Your question is answered in the article on Kepler's_laws#Acceleration_vector. The 'perpendicular' velocity is

\scriptstyle V_{p}=r{\dot {\theta }}

and its rate of change is

\scriptstyle {\frac {dV_{p}}{dt}}=r{\ddot {\theta }}+{\dot {r}}{\dot {\theta }}

while the perpendicular acceleration

\scriptstyle r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}={\frac {d(rV_{p})}{rdt}}

, including the extra coriolis term

\scriptstyle {\dot {r}}{\dot {\theta }}

, is zero for central force. It is tricky stuff. Bo Jacoby (talk) 11:10, 9 June 2010 (UTC).[reply]

Sorry for disrupting Wikipedia. I am not overturning Newton,(this is your mis-understanding) but Kepler's laws with the laws of Newton. I say: Newton's laws do not confirme Kepler's laws. That is all.(Is this an overturning of Newton's laws ?). And for this I ask Wiki people to not to copy -paste the old mathematicians math but make the derivations themselves. For example: you mean d(Vp)/dt=0. You write d(r*Vp)/(dt*r)=0.It is not you who write this, you just copy-paste the article (Kepler's_laws#Acceleration_vector). I don't want to consider this as an answer. According me, this is to multiply the inside of the parenthesis with (r) and divide the outside of the parenthesis with (r). So, you say nothing has changed, as you multiply and divide with the same value (r). Then by integrating, you deduct (r*Vp=Constant). Tricky math, hein! Multiply the interior of the parenthesis with (r^2) and then divide the exterior of the parenthesis by (r^2). And deduct (r^2*Vp=Constant). You discover the "Volume's law of Wiki"!!. Good job. TASDELEN (talk) 15:52, 9 June 2010 (UTC)[reply]

I have derived Kepler's laws from Newton's from scratch multiple times as, I expect, have the other people trying to explain this to you. It works. I'm not going to waste my time trying to explain exactly where you've gone wrong on each of the many mistakes you have made when you clearly don't actually want to learn. --Tango (talk) 16:55, 9 June 2010 (UTC)[reply]

Last chance: Is d(Vp)/dt=0 according (Kepler's_laws#Acceleration_vector)? Yes or no. TASDELEN (talk) 17:31, 9 June 2010 (UTC)[reply]

Ok, I'll give you one last chance: No, I don't think so (assuming I'm correctly understanding your notation - the article doesn't use Vp, but I'm assuming it is the magnitude of the perpendicular component of velocity). The perpendicular acceleration is zero, but that acceleration isn't simply the change in the perpendicular velocity, it also has a term that results from the changing angle.

V_{p}=r{\dot {\theta }}

so

{\frac {\mathrm {d} V_{p}}{\mathrm {d} t}}=r{\ddot {\theta }}+{\dot {r}}{\dot {\theta }}

. The perpendicular acceleration, however, is

r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}

(see Kepler's_laws#Acceleration_vector for derivation - the extra term comes from differentiating the radial velocity,

{\dot {r}}{\hat {\mathbf {r} }}

, and the product rule gives us a term with

{\frac {\mathrm {d} {\hat {\mathbf {r} }}}{\mathrm {d} t}}

in it, which is a perpendicular term). It is that perpendicular acceleration that is zero. --Tango (talk) 18:14, 9 June 2010 (UTC)[reply]

(edit conflict). The answer is no!

\scriptstyle {\frac {dV_{p}}{dt}}

is not zero in a noncircular Kepler orbit. And control your attitude regarding last chance. You know the rule for differentiating a product: d(x·y)=dx·y+x·dy. Then what is

\scriptstyle {\frac {dV_{p}}{dt}}={\frac {d(r{\dot {\theta }})}{dt}}

? Is it equal to

\scriptstyle {\frac {d(rV_{p})}{rdt}}={\frac {d(r^{2}{\dot {\theta }})}{rdt}}

? Yes or no. Bo Jacoby (talk) 18:30, 9 June 2010 (UTC).[reply]

Okey.Your answer for d(Vp)/dt=0 is NO. You say d(r*Vp)/(r*dt)=0. Then you deduct (r*Vp=Constant). I say your statement may be correct, but the integration is wrong. So, the deduction is wrong. According me the derivation of your statement should be as follows: [model d(y/x)=(dy*x-dx*y)/x^2]

[d(r*Vp)/(r*dt)]'= [(r'*Vp+r*Vp')*r-r'*(r*Vp)]/r^2= Vp'= 0

When integrating Vp=Constant ????!!!!!......Forget it.

Anyhow,we must proceed with the following questions: (copy from "Is the following article removed?")

10- If (1/2*m*Vr^2+m*g*r+1/2*I*w^2=m*r*dVr) the energy conservation equation rigth ,( Vp doesn't matter) then

11-Is 1/2*m*(dr/dt)^2+K/dt^2=m*r*d(dr/dt)/dt where (K/dt^2=m*g*r+1/2*I*w^2)rigth? (with g variable) If yes,

12-Is r'^2+2*K/m/dt^2=2*r*r" the differential equation rigth? (all terms are of the same order), If yes,

12 alternative- You may write your differential equation.

13-Is r=-a*t^2+a*t*tmax+K where K=-a*tmax^2/4=Constant and this solution is rigth? If yes

14-How to comment r=-a*t*(t-tmax) graph on polar plane coordinates? (Even on Cartesians)

15-Does the last equation indicate the orbital motion of the planets? (3 dimension). TASDELEN (talk) 19:40, 6 June 2010 (UTC). TASDELEN (talk) 11:37, 10 June 2010 (UTC)[reply]

I said that was your last chance. I'm not going to go through all your other mistakes explaining them in detail. You seem to be differentiating when you should be integrating, though... --Tango (talk) 14:46, 10 June 2010 (UTC)[reply]

Hasn't Meni explained the errors sufficiently on the Maths Ref Desk? Dbfirs 14:58, 10 June 2010 (UTC)[reply]

Your statement [d(r*Vp)/(r*dt)=0], when differentiated, gives Vp'=0. That is Vp=Constant. I don't mean integrating. You are integrating [d(r*Vp)/(r*dt)=0], and say [r*Vp=Constant]....Forget it, assume you are rigth. Proceed with the other questions, please. Thanks. TASDELEN (talk) 18:11, 10 June 2010 (UTC)[reply]

You might be able to prove it by differentiating (I haven't tried), but it is far easier to do it by integrating. Multiply both sides by r, integrate wrt t, divide both sides by r, and you're there - Vp=constant. I'm not answering your other questions. Learn the relevant maths and do it yourself - it's very simple stuff. --Tango (talk) 22:37, 10 June 2010 (UTC)[reply]

TanksTango.I posted the following comments on "Reference desk mathematics".So,you may close this discussion as "Resolved".

It seems,I disturbed you enormly. Sorry. Please agree my apologizings. My intention was not to disturb your believings. I believe to the laws of Newton. I considered F*dt*r=m*dv*r , wrote its differential, discovered r=-a*t*(t-tmax)+K. That was the equation of the planetary motion, according me. That was against Kepler's laws, said to be confirmed by Newton (or attributed to Newton by later mathematicians). Two different solutions using Newton's laws; one of the solution should be wrong. I just wanted to share my reasoning with you. You considered it as a lecture. And refused to comment on this equation of motion. According Wiki's canons you are rigth. Forget please all I wrote you. I feel you started to think on this matter. Thanks88.234.182.137 (talk) 20:23, 11 June 2010 (UTC).This is TASDELEN (talk).Automatic sign is wrong.TASDELEN (talk) 20:39, 11 June 2010 (UTC)[reply]