Talk:Wheelie

Getting up or not

"To perform a wheelie on a bicycle, put the bike in low gear, get up from the saddle and pedal forward while pulling hard on the handlebars."
I can do a wheelie just fine without getting up from the saddle. The description should be changed. κаллэмакс 22:31, 19 June 2006 (UTC)

I agree. Friends of mine can cycle for long distances on one wheel - and they all do it siting down. I'll change it.Adrian Baker 12:04, 26 June 2006 (UTC)

Removed a line saying that you can only pull a wheelie from a starting position, and that if you do it while moving you will not succeed. Seeing as I have been able to pull "successful" wheelies on my bike at 20+ mph, and you see people do them while moving forward all the time on TV or at your local park, it's obviously wrong. Replaced it with a line qualifying that it's _easier_ to do it while stopped, and experts can do it while moving. 67.158.111.66 19:02, 21 August 2007 (UTC)

The faster you are traveling when you pull a wheelie the more stable it will be because of the gyroscopic effect of the wheels spinning. It is safer to do a wheelie at a high speed than a low speed as almost no gyroscopic effect will allow you to tip to either side and not maintain a straight line as easily. —Preceding unsigned comment added by 66.251.89.139 (talk) 15:36, 10 March 2011 (UTC)

isnt it true that the air only makes auto wheelys easier since the force is in the same way as the force of accelerating the bike. is there any explanation of the formula? — Preceding unsigned comment added by 94.209.180.23 (talk) 23:11, 10 January 2014 (UTC)

That depends on what you mean by "easier". If you mean the power required, the effect of air drag will depend upon the height of the center of pressure relative to the height of the center of mass. Below is a simplistic analysis to demonstrate this point:

File:Top fuel dragster with air drag.pdf.png

This might make a nice addition to the article, if anyone can find a suitable reliable source for it. Otherwise, it might be mistaken for original research. -AndrewDressel (talk) 19:12, 11 January 2014 (UTC)

Wheelies illegal on the street?

It appears wheelies bring a misdemeanor charge, reckless driving, if done on the street (at least in virginia) [23]. Anyone know if there is any sort of federal traffic law against it, or does it vary by state, or even county? --Fxer 17:39, 3 August 2006 (UTC)

Most state statutes have a clause in the speeding laws that allow a officer to write up a ticket / pull you over to talk / for accelerating at a unreasonable speed as it is called a : Exhibition of speed : which can also be: Reckless : and that is illegal in the USA and allows the officer a gray area where he has a list of things he can charge you with depending on how wild your maneuvers were and unfortunately a large part of your ticket comes down to what kind of mood the officer is in at the time and if he has something better to do or not.

"A wheelie is a willful disregard of safety (aka, Reckless Operation) I issue Reckless driving tickets to those who pull wheelies. Public roads are not the place for your stunt driving and I am tired of responding to motorcycle down collisions." Statement : made from local law officer in Utica NY

I know a friend of mine talked his ticket down to an "unsafe start" because of an accidental wheelie pulling away from lights. —Preceding unsigned comment added by 66.251.89.139 (talk) 15:19, 10 March 2011 (UTC)

Copyvio

I removed a large chunk of text added by User:Burrelly (contribs), because it was stolen from this site: [24]. If you disagree with my revert, discuss here. - Tronno ( t | c ) 14:09, 18 April 2007 (UTC)

First image (RC car)

If that vehicle is still jumping, I think then it would not count as a wheelie. As soon as it touches the ground, it will come down on all four wheels. Should be removed, especially as we have better images, and this one doesn't even show it all that well from a purely technical point. Ingolfson 10:03, 3 August 2007 (UTC)

Deletion

This article is being considered for deletion so I added some wheelie history. Perhaps the focus of the page should turn to such topics as wheelie records and accomplishments, which I will add a section about. —Preceding unsigned comment added by 128.205.15.127 (talk) 04:30, 18 December 2009 (UTC)

Do you have any sources for this? Adding more unsourced information doesn't actually help.--Dbratland (talk) 05:13, 18 December 2009 (UTC)

Yes I posted some sources. —Preceding unsigned comment added by 128.205.190.26 (talk) 21:03, 20 December 2009 (UTC)

You should probably review Wikipedia:Reliable sources. Blogs and Wikis and such anonymous sources don't cut it; it needs to be the kind of source where there is some sort personal accountability. Where you know who wrote it, and you know that person is a respected source, and hopefully has some sort of editing oversight or peer reviw. Such as a published book, magazine, newspaper, etc, or a web page authored by a named individual recognized by others as an authority.--Dbratland (talk) 22:59, 20 December 2009 (UTC)

• Also! Please Create an account.--Dbratland (talk) 23:02, 20 December 2009 (UTC)

Why is a citation needed for wheelies being used to get over a log in the first place? If you ride straight into a sandy patch the front wheel will get stuck in it and drop down in it, what citation could possibly be needed? It is listing an obvious use of the maneuver. —Preceding unsigned comment added by 128.205.190.25 (talk) 20:01, 21 December 2009 (UTC)

Is info about legal status valid content for this article?

72.82.227.99 (talk · contribs) has twice deleted the legal status information in the motorcycle section - most recently today, arguing that it adds nothing to the article. I disagree and have added other citations to show that it is not just the UK where wheelies on a public road is illegal. What do others think? Is it relevant?--CIHAGM (talk) 13:43, 25 January 2010 (UTC)

• It seems relevant to me, especially in light of the delete discussion in which this article was described as merely a dictionary entry and a how-to. Other, well-sourced, aspects of wheelies should also be welcome. -AndrewDressel (talk) 14:03, 25 January 2010 (UTC)

File:Indian wheelie.JPG Nominated for speedy Deletion

An image used in this article, File:Indian wheelie.JPG, has been nominated for speedy deletion for the following reason: All Wikipedia files with unknown copyright status What should I do? Don't panic; you should have time to contest the deletion (although please review deletion guidelines before doing so). The best way to contest this form of deletion is by posting on the image talk page. If the image is non-free then you may need to provide a fair use rationale If the image isn't freely licensed and there is no fair use rationale, then it cannot be uploaded or used. If the image has already been deleted you may want to try Deletion Review This notification is provided by a Bot --CommonsNotificationBot (talk) 14:52, 27 September 2011 (UTC)

Fastest motorcycle wheelie

The 1999 Furstenhoff record seems to be out of date. According to Motor Cycle News, he did a 215 mph wheelie in 2006.[25] — Brianhe (talk) 06:48, 4 September 2014 (UTC)

No reference to "Torque Wheelie"

Torque Wheelies occur when the engine's torque is sufficient to twist the vehicle along its length to the point where one of the front wheels raises off the ground while the other doesn't. It occurs most often in Funny Cars but does occur occasionally with Top Fuel Dragsters. TCav (talk) 16:03, 28 February 2016 (UTC)

Wheelie with a wheelchair

It's very important to know how to do a wheelie with a wheelchair. Some information about this would be welcome. Pe-ga-sos (talk) 13:53, 23 October 2016 (UTC)

It helps the user to get over higher obstacles than a doorstep, such as pavings. So it's more than just a dance move. Pe-ga-sos (talk) 15:23, 23 October 2016 (UTC)

It is done by leaning back and and simultaneously propelling the rear wheels forward. Pe-ga-sos (talk) 15:37, 23 October 2016 (UTC)

Now there's something. Check the grammar! Pe-ga-sos (talk) 17:03, 23 October 2016 (UTC)

New material concerning "factor of power translations"

I am moving the recent addition and subsequent comments in the edit summary here so that the discussion can be commenced properly. Please do not restore it to the article until we can resolve this conflict. -AndrewDressel (talk) 13:54, 16 November 2018 (UTC)

There is also a factor of power translations (M) which considers that the total engine power of a vehicle is divided into two work streams. The first being the power utilization in maneuvering and secondly being the power consumed in undergoing a wheelie.The value of M can give us the idea of if the vehicle under consideration will undergo wheelie or not. For those who want to desirably stimulate a wheel-stand, like professional stunt drivers, should make sure that value of M remains close to 1. On the other hand, those who want to avoid a wheelie for the sake of handling, like drag racers, should make sure that center of mass of the vehicle is away from the rear wheel to prevent wheel-stand under maximum acceleration

$M_{}=g\pi xn{\frac {[\theta .\sin \theta +\cos \theta -1]}{av_{i}}}$

where $g$ is the Acceleration due to gravity, $x$ is the longitudinal distance between point of center of gravity and rear axle, $\theta$ is the Angle of wheelie in degrees with respect to road surface, $a$ is the acceleration of vehicle, and $v_{i}$ is the velocity of vehicle at any point of time. .

"I added an additional formula for factor of power translations, it can give us the idea of if the vehicle under consideration will undergo wheelie or not. For those who want to desirably stimulate a wheel-stand, like professional stunt drivers, should make sure that value of M remains close to 1" - 12:50, 16 November 2018‎ Gursagar virdi (talk | contribs)

Which I removed with the following edit summary "Rv addition by same author of paper used for referenced: COI." - 17:52, 16 November 2018‎ AndrewDressel (talk | contribs)‎

Which the original author re-added with the following edit summary "The previous edit is illogical as the removed content regarding "Power Translation Factor" is solely different from the used references and holds a genuine proof of intellectual advances as compared to mentioned references." - 23:05, 16 November 2018‎ Gursagar virdi (talk | contribs)‎

Well, it appears that one of two cases must be true:

1. Either the cited source, written by Gursagar Singh, does support the new material, written by Gursagar virdi, in which case there is a conflict of interest and the new material does not belong in the article, per WP:SELFCITE, or
2. The cited source does not support the new material, in which case the new material is original research and does not belong in the article, per wp:or.

Which is it? - AndrewDressel (talk) 13:54, 16 November 2018 (UTC)

If somehow neither case applies, there are several additional issues with the new material that should be resolved:

1. The cited source that provides the equation does not include n, which is also not described in the contested text. What does n represent?
2. It is not clear what either Gursagar mean by θ.cosθ. Is that supposed to mean the product of θ and cosθ? I haven't see the period symbol, ".", used that way before. Why use a symbol at all when none is used for any of the other implied multiplications?
3. The "velocity of vehicle at any point of time" is represented by $v_{i}$, but no other measurement we expect to change in time, angle θ and acceleration a, has subscript i. Why just velocity?
4. Finally, the given expression for M seems to suggest that if a vehicle is not already in a wheelie, it can never enter one. If it is not in a wheelie, then θ = 0, sinθ = 0, and cosθ - 1 = 0, so sinθ + cosθ - 1 = 0, and M = 0. The author asserts, however, that "those who want to desirably stimulate a wheel-stand ... should make sure that value of M remains close to 1." How does that work?

The cited paper appears to have further problems, but I'll have to leave that for another time. -AndrewDressel (talk) 14:27, 16 November 2018 (UTC)

Hello Andrew, Your revision to my edit in wheelie is very unfortunate can't understand why you are doing so. Since its a published research paper you cant ignore the fact that it is genuine in every sense. I will keep reverting your revision until you stop or prove the formula is not genuine. Thank you. A Researcher. — Preceding unsigned comment added by Gursagar virdi (talk • contribs) 16:18, 16 November 2018 (UTC)

There's a lot in there, so allow me to respond with a numbered list:

1. I have clearly explained why I am removed the content each time I did so. The editor adding the content appears to be the author of the cited paper, and thus the author has a conflict of interest.
2. You are certainly welcome to keep reverting things until you get blocked from editing Wikipedia for disruptive editing.
3. I don't have to prove that the formula is not genuine. I merely need to convince other editors that there is a conflict of interest and point out that there are other serious flaws with the content, as I have already enumerated above.
4. No, there is no indication that the research paper is "genuine in every sense". It appears to be published in the proceedings of "2018 3rd International Conference on Design and Manufacturing Engineering (ICDME 2018)", which doesn't appear to have any expertise in vehicle dynamics. It isn't clear that there is any review of the paper at all. In fact, it appears to have serious flaws:

• After the height of the center of mass, $y_{1}$, is calculated in section 2.2, it does not appear to be used anywhere. It is already well established in the article that the power required to perform a wheelie depends on both the horizontal and the vertical distance from the rear wheel to the center of mass. After all, if the vertical distance is zero, then no amount of power will cause a wheelie.
• In (Eq.10) at the top of page 4, two different instances of the same symbol $\theta$ are used to represent two different things: the angle of the bike relative to the horizontal from $x_{2}=\cos \theta$ in figure 7, and the rate of change of the angle of the bike from $\omega =(\pi /180)\theta$ °/sec (Eq.8), in which it should be written $d\theta /dt$ or ${\dot {\theta }}$. These are two distinctly different things and simply cannot be represented by the same symbol.
• How does (Eq.10) $P_{1}=mgx_{2}\cos \theta ({\frac {\pi }{180}})\theta$ become $\int _{0}^{\theta }{\frac {P1}{d\theta }}=\int _{0}^{\theta }[mgx_{2}\cos \theta ({\frac {\pi }{180}})\theta ]d\theta$ in the next paragraph? From where do the $d\theta$ in the denominator of the left side and the $d\theta$ in the numerator of the right side magically come? I could see multiplying both the left and right side of (Eq.10) by $d\theta$, but then it would appear in the numerator on both sides. Is it not equivalent to writing $y=7x\Rightarrow \int {\frac {dy}{dx}}=\int 7xdx\Rightarrow y={\frac {7}{2}}x^{2}$, which is nonsense.

I look forward to a response to any of the issues I have raised. -AndrewDressel (talk) 19:33, 16 November 2018 (UTC)

I have seen all your so called allegations of my paper not being genuine and flawful. Let me tell you Mr. Andrew the paper has been peer reviewed by a council in Monash university, Melbourne before being accepted for publication. The ICDME2018 published 6 papers on vehicle dynamics along with mine and the Y2 vertical distance about which you were talking about is actually Y1 (get your eyes checked, No Y2 exists) and it doesn't play any role in equation of power utilization. Its X2, the horizontal distance of center of mass from rear wheel which is contributing in Power utilization not the vertical distance. In indian terminology of integration D/thetha is used in denominator and in numerator to express mathematical equilibrium but are not taken to consideration. Please get your facts checked along with your eye sight Mr. Andrew, Seems like you need to go to high school again. — Preceding unsigned comment added by Gursagar virdi (talk • contribs) 04:13, 17 November 2018 (UTC)

Please remain civil and keep your comments about my eyes to yourself. Whether it is y1 or y2 does not matter. What does matter is that your paper spends time deriving it and then does not use it, and the vertical distance to the center of mass is crucial to understanding wheelies. Meanwhile, you have not yet addressed most of the other issues I have raised:

1. Citing your own paper is a clear conflict of interest, and that is what has let to all this scrutiny of it.
2. Your don't consider the height of the center of mass in the performance of a wheelie, despite spending nearly a full page of the paper in an attempt to derive it, and it is already established as a crucial parameter. A wheelie is impossible if the center of mass does not have a finite height above the pavement.
3. You incorrectly equate $\omega$ to $\theta$ instead of the correct $d\theta /dt$ or ${\dot {\theta }}$.
4. Your calculus makes no sense and "in indian terminology of integration D/thetha is used in denominator and in numerator to express mathematical equilibrium but are not taken to consideration" does not clear anything up because "D/thetha" does not appear anywhere in your derivation. Perhaps you mean $d\theta$, the differential of the variable $\theta$, used to indicate that the variable of integration is $\theta$. But if that is what you mean, then it can't "express mathematical equilibrium." Perhaps you can provide a link to some article that explains "indian terminology of integration."
5. Finally, as already stated above, the given expression for $M$ seems to suggest that if a vehicle is not already in a wheelie, it can never enter one, at least for how $\theta$ is defined in Figure 3 and Figure 4(a). If it is not in a wheelie, then $\theta =0$, $\sin \theta =0$, and $\cos \theta =1$, so $\theta \sin \theta +\cos \theta -1=0$, and so $M=0$ for any values of $g$, $x$, $a$, and $v$. The author asserts, however, that "those who want to desirably stimulate a wheel-stand ... should make sure that value of M remains close to 1," and "if M<1: vehicle will not undergo a wheelie." So, how can this possibly work?

When you are ready to skip the personal attacks and respond to these issues, I look forward to your response. -AndrewDressel (talk) 05:41, 17 November 2018 (UTC)

Meanwhile, the contested content should remain here until we can get some answers to all these questions. I've already tried to clean up the easy errors, but the fundamental flaws remain unaddressed. -AndrewDressel (talk) 16:08, 17 November 2018 (UTC)

The following is my response--Gursagar virdi (talk)

1. Citing my own paper will only be considered conflict of interest if the added content is irrelevant to page, Thus this doesn't applies on me. -Gursagar virdi (talk)

That is not how Wikipedia works. Per wp:SELFCITE "Citations ... should not place undue emphasis on your work. When in doubt, defer to the community's opinion: propose the edit on the article's talk page and allow others to review it." -AndrewDressel (talk) 19:15, 17 November 2018 (UTC)

2. As shown in Figure 7 of the research paper X2Cos $\theta$ is a function of vertical distance thus the final equation does considers the effect of vertical distance. -Gursagar virdi (talk)

No. This is not what I mean by vertical distance. Instead, the vertical distance that must be considered is measured perpendicularly from the line between the front and rear contact patches, or the ground plane if a wheelie has not yet occurred, and the wheelie angle $\theta$ is zero. This is the definition already used throughout the article. The expression you cite, $x_{2}\cos \theta$, will be zero when the wheelie angle $\theta$ is zero, and if the perpendicular distance from the ground plane to the center of mass is zero, a wheelie can never happen. Instead, the correct and complete expression for the vertical distance from the ground to the center of mass, using your notation, is $x_{2}\sin \theta +y_{1}\cos \theta$, the horizontal distance is $x_{2}\cos \theta -y_{1}\sin \theta$, and the total distance is ${\sqrt {x_{2}^{2}+y_{1}^{2}}}$. -AndrewDressel (talk) 19:15, 17 November 2018 (UTC)

3. "You incorrectly equate $\omega$ to $\theta$ instead of the correct $d\theta /dt$ or ${\dot {\theta }}$." Please mention the equation number. -Gursagar virdi (talk)

(Eq.8), in which you write $\omega =(\pi /180)\theta ^{\circ }/\sec$. You are merely converting radians to degrees and complete ignoring the fact that $\omega ={\frac {d\theta }{dt}}={\dot {\theta }}\neq \theta$. This has important implications for the integration you attempt to perform later in the paper. -AndrewDressel (talk) 19:15, 17 November 2018 (UTC)

4. This doesn't matter because final results of integration doesn't change. If not, You can let me know the correct equation if you get different results. -Gursagar virdi (talk)

This is pretty meaningless, because there are flaws in the derivation of (Eq.10), but this is how I would perform the integration correctly:

• $P_{1}=mgx_{2}\cos \theta ({\frac {\pi }{180}}){\frac {d\theta }{dt}}$ - correct (Eq.10) to account for $\omega ={\frac {d\theta }{dt}}$
• $P_{1}dt=mgx_{2}\cos \theta ({\frac {\pi }{180}})d\theta$ - separate variables to solve differential equation
• $\int _{0}^{t}P_{1}dt=\int _{0}^{\theta }mgx_{2}\cos \theta ({\frac {\pi }{180}})d\theta$ - setup definite integral
• $P_{1}t=mgx_{2}(\theta \cos \theta +\sin \theta ){\frac {\pi }{180}}$ - perform integration, assuming $P_{1}$ is constant in time
• $P_{1}=mgx_{2}{\frac {(\theta \cos \theta +\sin \theta )}{t}}{\frac {\pi }{180}}$ - isolate $P_{1}$

I have no idea what you might do with this corrected result because I am not trying to make this analysis, just pointing out and correcting mistakes, but at least the units are correct. Now the calculated power has SI units of $\mathrm {kg} \cdot {\frac {\mathrm {m} }{\mathrm {s} ^{2}}}\cdot {\frac {\mathrm {m} }{\mathrm {s} }}=\mathrm {kg} \cdot \mathrm {m} ^{2}\cdot \mathrm {s} ^{-3}=\mathrm {W}$. That will also allow your factor M to be unitless instead of the units of time that it currently has: ${\frac {x}{v}}=t$. -AndrewDressel (talk) 19:15, 17 November 2018 (UTC)

5. No, your analytics are week. Please read point "g" of Conclusions section and overall article tries to convey that the vehicle can only undergo a wheelie if is able to attain the needed acceleration. -Gursagar virdi (talk)

I have already read and quoted point "g", specifically

"g) ... The value of M can give us the idea of if the vehicle under consideration will undergo wheelie or not; ... 2. If M<1: Vehicle will not undergo a wheelie."

There is no mention of attaining "the needed acceleration".

The problem, besides the calculus issues, is likely that your expression for M does not incorporate the height of the center of mass, which I have defined above to be "measured perpendicularly from the line between the front and rear contact patches, or the ground plane if a wheelie has not yet occurred, and the wheelie angle $\theta$ is zero." -AndrewDressel (talk) 19:15, 17 November 2018 (UTC)

And, I now see that you again have not waited until these issues are resolved before restoring the disputed content. That is disappointing. -AndrewDressel (talk) 19:15, 17 November 2018 (UTC)

In any case, merely out of curiosity, here is what I believe to be a better derivation of an expression for the power required to perform a wheelie at some acceleration $a$, some velocity $v$, some angle $\theta$, with some angular acceleration ${\ddot {\theta }}=\alpha$, some angular velocity ${\dot {\theta }}=\omega$, with the center of mass $m$ at $x_{2}$ and $y_{1}$, and with a mass moment of inertia $I_{z}$, following the derivation presented near the top of this page, and using the free body diagram on which it is based, but accounting for non-zero $\theta$ and ${\ddot {\theta }}$ and neglecting air drag and limitations on available friction:

1. $\sum _{}^{}F_{x}=f=ma_{x}$ : linear momentum balance in horizontal direction with friction force $f$ at rear tire
2. $\sum _{}^{}F_{y}=N-mg=ma_{y}$ : linear momentum balance in vertical direction with normal force $N$ at rear tire
3. $\sum _{}^{}M_{/G}=f(x_{2}\sin \theta +y_{1}\cos \theta )-N(x_{2}\cos \theta -y_{1}\sin \theta )=I_{z}\alpha$ : angular momentum balance about center of mass $G$ with moments at rear tire from from friction force $f$ and normal force $N$
4. $P_{T}={\vec {F}}\cdot {\vec {v}}+M\omega$ : power is force dot velocity + moment times angular velocity, in 2D (Beer and Johnston equations 13.13 and 17.13)
5. $a_{y}=(x_{2}\cos \theta -y_{1}\sin \theta )\alpha$ : kinematic relationship between $a$, $r$, and $\alpha$
6. $I_{z}=m\cdot k^{2}$ : relationship between mass moment of inertia $I$ and radius of gyration $k$
7. $P_{T}=ma_{x}v_{x}+m(x_{2}\cos \theta -y_{1}\sin \theta )\alpha \omega +m\cdot k^{2}\alpha \omega$ : combine to eliminate $f$, $N$, $I_{z}$, $a_{y}$, and $g$ for vehicle with mass $m$ and dimensions $x_{2}$ and $y_{1}$ at instant $a=a$, $v=v$, $\theta =\theta$, $\omega =\omega$, and $\alpha =\alpha$ (units are correct $\mathrm {kg} \cdot \mathrm {m} ^{2}\cdot \mathrm {s} ^{-3}=\mathrm {W}$)
8. $P_{T}=m[a_{x}v_{x}+(x_{2}\cos \theta -y_{1}\sin \theta +k^{2})\alpha \omega ]$ : regroup to simplify slightly
9. $\alpha ={\frac {a_{x}(x_{2}\sin \theta +y_{1}\cos \theta )+g(x_{2}\cos \theta -y_{1}\sin \theta )}{k^{2}+(x_{2}\cos \theta -y_{1}\sin \theta )^{2}}}$ : can also express $\alpha$ in terms of accelerations $a_{x}$ and $g$ and vehicle dimensions $x_{2}$, $y_{1}$, and $k$, but it only make expression of $P_{T}$ a lot bigger.

So that reminds me that your expression for power also does not incorporate angular acceleration nor mass moment of inertia of the vehicle. -AndrewDressel (talk) 21:29, 17 November 2018 (UTC)

Finally, now that I'm reading your conclusion more closely, "rollover" is a term already used in vehicle dynamics to mean a vehicle tipping "over onto its side or roof," and your paper concerns rotation about a different axis. -AndrewDressel (talk) 21:29, 17 November 2018 (UTC)

Thank You Andrew for you inputs. You have been right. I followed your feedback and compared it. you seem right but its a minor mistake which is making a big difference afterwards. Our goal is to make wikipedia more knowledgeable and I'm glad you rectified my formula, I have updated it according to your calculations. For the sake of giving correct knowledge to wiki users, I request you to please don't revert the recent edit done by me. I know you will say that doesn't match up with reference hence is not viable to be projected on page. it would be good if you could find a way to cite your calculations here without removing my cite. Thank you -Gursagar virdi (talk) —Preceding undated comment added 14:09, 18 November 2018 (UTC)

While I am still not sure of your/our conclusion, the article no longer contains the errors I pointed out above. What you are suggesting, including a result we derive here on the talk page, is strictly against Wikipedia's prohibition against original research, but I would really like to stop the slow edit war we've been having, and there appears to be no other editor(s) to end the stalemate. I'll leave your recent change, after correcting the typo, for now, but if and when another editor shows up to complain about our method, I will have to agree with them. In the meantime, you may want to start working on publishing a correction to your paper. -AndrewDressel (talk) 15:16, 18 November 2018 (UTC)

Also, now that there is a new expression for M, is there any reason to believe your original assertions about what it's values mean? Does M = 1 have any signifigance anymore? I haven't checked. Have you? -AndrewDressel (talk) 15:39, 18 November 2018 (UTC)

Physics

"Since mechanical power can be defined as force times velocity, in one dimension, and force is equivalent to mass times acceleration, then the minimum power required for a wheelie can be expressed as the product of mass, velocity, and the minimum acceleration required for a wheelie:

Pmin=mvamin " The formula passes unit analysis, but still feels suspicious. As said in the article, the force and velocity need to be in one dimension - it seems to me the speed at which the system is moving relative to the ground should not be relevant to the minimum power required. The formula doesn't have any sources either, so could someone check this? BlueBanana (talk) 23:30, 21 February 2021 (UTC)