http://home.attbi.com/~bernhard36/honda-ad.html
yeah, yeah, yeah. It's a honda commercial, but you will enjoy it. trust me.
Read this after:
http://www.dailytelegraph.co.uk/news/main.jhtml?xml=/news/2003/04/13/nhonda13.xml&sSheet=/news/2003/04/13/ixhome.html
Thats sweet
Very Cool
WOW!!
Laughed my head off!!!
Geoff
Absolutely fascinating!
I can't believe they were able to talk them into doing it in real time!
It somehow gives me hope for mankind.....
Rube Goldberg would be astounded.
Excellent!!!!!!
CCLinLOO
First the S2000 and now this. Honda is working hard to get off my "makes I'd never own" list.
It's too bad we can't see the whole ad. The Honda UK web server is not responding.
I've saw it on TV during the Brazillian F1 GP. We were impressed.
We may not have rust free cars, but we have Honda ads!
I'd buy a Honda bike in a minute; and I will be the first to say that my anti-Honda stance is largely irrational and unjustified (largely, but not entirely).
I'm the only one here with 2 honda rice rockets i think.
The only "new" car I ever owned was an Acura Integra. In the 5 years I owned that car, I had to replace one headlight bulb, and that was covered under warranty service.
That, and the Acura dealership were VERY nice.
I'd recommend any Honda or Acura for that purpose, small family car. They're both a bit pricy, but for what you get, definately worth it, for the reliability.
Sporty? Well, the drive wheels are on the wrong end for one thing, so whatchya gonna do?
I was very happy when Renault finally kicked Honda's butt in F1 in 1992.
Has anyone gotten this ad to play all the way through? I keep getting a time out from the site and have never been able to get through to the original Honda Uk link.
I just saw that add the other day..... I think its really cool.
Hondas a great cars, not Porsches but great daily drivers.. My bros and I just bought my mom a 2003 accord for her bday.. I know for the next few years mom wont have a hint of car trouble..
WAY COOL
i like the whole spinning exhaust on the crank- cool pitch to it!
No way this is actually filmed (like the article suggests)
I have a serious problem with the wheels running up the ramp
And you'll see that the background is constantly the same repetition (just different lighting)
Still a very cool ad!
cheers,
Jeroen
I think they did this video on a slanted floor. Mainly because the wheels going up hill and something else that looked fishy.
Did you guys read the article? They said the wheels were weighted on the inside of the tire so all they needed was the little nudge to roll uphill.
I'm with Jeroen. That muffler or cat sure rolled alot of revolutions. And I couldn't stop the thing in the right places, but each tire would knock the next 1/2 the distance of travel, not to mention the inertia lost due to friction of the rubber meeting while travleing in opposite directions.
Jeroen and I are just a couple of spoil sports. <_<
Nice.
All in all a very cool commercial, and I could see how the tires rolled uphill if they were weighted on the inside, but, I have to agree with Milt. That muffler rolls about 2 revolutions too many.
I could only watch a few seconds of the ad but I don't see how anything could roll uphill starting from a stop no matter how it's weighted. As I understand it, the only way a rolling tire can roll uphill is by converting its kinetic energy into potential energy and if it has no kinetic energy to start, it's not going to start rolling uphill no matter how it's weighted. The center of mass of the tire is going to try to get to the place of least potential energy.
Jeff.. you're exactly right.. but thats exactly why the tires apppear to roll uphill. If a large point mass (larger than the wheel and tire itself) is initially at the top of the tire, it's highly unstable. the slightest bump will cause the tire to turn and lower the center of mass (and potential energy) of the system. It just looks funny 'cause you're used to the center of mass of a tire+wheel being at the hub. You'll notice none of the tires on the incline spin more that 180 degrees. (in fact they dont even make 180, if you think about it)
..sorry i didnt mean for that to turn into a mechanics lecture.
The tires rolling uphill is cool, and after reading the article and discussing it, we understand how it works. But, the casual TV viewer won't. The tires rolling uphill will discredit the entire ad. 99% of the viewers will say "cool but fake". :toilet:
Yeah, but what got that muffler to roll so many times?
ok, time to dig out my physics degree and explain how the tire thing works. (Alfred (or jeff k. as he calls himself these days) will love this!).
1. is the ramp
2. is the tire
3. is the weight inside the tire. important, the weight has to be more or eaqual to the weight of the tire itself. i'll come back to that in just a sec.
4. the force vector for the weight, going straight down.
5. the force vector for the tire. trying to roll down the ramp. (also tries to go down, but we don't want to make it too complicated, do we)
6. the lenght of which the tire is going to move "uphill"
7. the point where it will come to rest
and here is how it works:
first, you need to get away from the obvious, a tire moving up a ramp on his own. you need to look behind that and "visualize" what is actually going on. (Alfred, still with me?).
to prefectly balance the tire on the ramp, the weight has to be placed with an offset to the tire's TDC that is eaqual to the angle of the ramp. (actually, not exactly because we also have some friction of the tire to the ramp). this will hold the tire in place. if the weight was less than the weight of the tire, the tire would be able to "spin" the weight around the TDC and the tire would roll down or the placement of the weight would have to be so far to the right that the amount traveled upwards would be insignificant.
now any additional force (like something hitting from the left) will make the setup unstable. the weight will start moving STRAIGHT DOWN. but because it is attached to the tire, this will result in a clockwise rotation of the tire. the weight actually still goes down, while the tire moves up on the ramp. the tire will stop moving when the weight comes as close to the ramp as possible (point 7). by then the tire will have traveled the distance of 6. up the ramp.
everybody still with me?
Andy
Attached image(s)
You guys gotta remember that this was for a commercial in the U.K. where they don't have any technology to do this with computers. Remember this is the same country that gave us Dr. Who.
(flame suit on)
Dr. WHo! LOL!
YET another great commercial.
http://www.adforum.com/affiliates/EvS/003/ad_detail2.asp?type=reel&ID=28797&TDI=VDez8str
Oops! This solution is wrong. Check out my later post for the correct solution.
***************************************
Ok, Andy, I'm no physicist so you can correct me where I'm wrong.
Let the tire sit just at the bottom of a ramp which has a positive slope. If
the tire's center of mass is a distance r from the tire's center and the
center of mass is located at the top dead center point on the tire, then
it's height (y) above ground is given by y = r*[1+cos(a)] + r*a*sinb. Here
"a" is the angle formed between the vertical axis and the tire's (point)
center of mass. 12 o clock ---> 0 rads and 3 oclock ---> pi/2 rads etc. "b"
is the angle that the ramp makes with the horizontal axis.
Since the tire's mass (m) and the acceleration due to gravity (g) are both
constant, the critical points of the tire's potential energy (mgh) equation
are determined completely by the
the height of the tire's center of mass above the ground h = y =
r*[1+cos(a)] + r*a*sinb. For fixed r and b, dy/da = r*sinb - r*sin(a). The
critical numbers occur when r*sinb = r*sin(a), when a = b or a = [pi - b].
When the ramp makes an angle of b with the ground, the first critical number
occurs at an angle of a = b. You can verify that a = b produces a local
maximum. Therefore the tire has its maximum potential energy when a = b and
as the tire rolls further up the ramp its PE decreases until the angle a =
[pi - b] when it reaches a local minimum.
Here is a graph for y = r*[1+cos(a)] + r*a*sinb for the specific case when r
= 10 and b = pi/3. (In this graph x = a).
You couldn't figure out my point, Andy? For the tire I described above, if you position it so that the center of mass makes an angle with the vertical equal to the angle [b] that the ramp makes with the ground and then give the tire a small uphill push, it will "roll" uphill slightly until it reaches the point of (relative) lowest potential energy at a = [pi - b] radians.
And, yeah, it's DOS based software from 1993 and it's pretty basic but it works well and takes the tediousness out of many math operations. If you want to help, I'm willing to take donations that go towards the purchase of Mathematica?
BTW, Andy, why no comment on the emails from the math Profs that I forwarded to you concerning the basic math question that I asked you earlier?
A man does not stand taller standing on someone else.
Everything doesn't have to be a contest of intelect. Can't you accept the fact that YOU ARE BOTH INTELLIGENT and STFU about it?
-Ben M. (still using the old student edition of Mathmatica made for win3.1)
i told him to keep it off-list, but he won't ...
oh great!
the arguments continue!
lets just leave it at..."hey man, what a cool commercial!"
PLEASE
On the other hand, they probably could have just saved us the physics lesson and done it with computer animation.
My previous solution to the "tire rolling uphill" question was incorrect.
Here's the correct (I hope) solution. If you can find a mistake or know of a
simpler solution, let me know.
Let the tire sit just at the bottom of a ramp which has a positive slope. If
the tire's center of mass is a distance r from the tire's center and the
center of mass is located (initially) at the origin of the x-y coordinate plane (see my
diagram below), it's height (y) above ground is given by y = r(a-sina)sinb +
r(1-cosa)cosb. Here "a" is the angle formed between the radius of the center
of mass and the angle -pi/2 + b. "b" is the angle that the ramp makes
with the horizontal axis.
Since the tire's mass (m) and the acceleration due to gravity (g) are both
constant, the critical points of the tire's potential energy (mgh) equation
are determined completely by the the height of the tire's center of mass
above the ground h = y = r(a-sina)sinb + r(1-cosa)cosb. For fixed r and b,
dy/da = r(sinb)-r(sinb)(cosa) + r(cosb)(sina).
For the specific case where r = 4 and b = pi/6, dy/da = 2sqrt(3)(sina) -
2(cosa) + 2 which has critical numbers a = 0 + 2n(pi) and a = 4/3(pi) +
2n(pi) where n is an integer. a = 0 corresponds to the tire's center of mass
being at the origin and having 0 potential energy, a = 4/3(pi) corresponds
to the tire's center of mass being pi/6 radians after TDC (see my diagram
below) and is a local maximum. If at this point the tire is given a slight
uphill push, then it will roll uphill until it reaches a = 2(pi) radians
which is a local minimum for the potential energy equation. The rotation of the tire from a = 4/3(pi) to a =
2(pi) radians moves the center of the tire 4(2 - 4/3)pi units up the hill.
Here is a graph for y = r(a-sina)sinb +
r(1-cosa)cosb for the specific case when r
= 4 and b = pi/6. (In this graph x = a).
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