Author |
Message |
Lovematt
| Posted on Thursday, April 07, 2005 - 08:36 am: |
|
Blake, As for when MSIE decides to overwrite or not...I have found the following settings in IE...don't know if they apply here: 1. Select Tools and then Internet Options. 2. In the General Tab click on Settings for Temporary internet files. 3. At the top there are settings to check for a newer version of a page each time, only when starting IE, automatically, and never. My guess is "never" would download the page one time and then never check/update it again... |
Ebear
| Posted on Thursday, April 07, 2005 - 10:09 am: |
|
THANKS BLAKE..!!!!I'm baaaacckkk.......!!!!
|
Outrider
| Posted on Thursday, April 07, 2005 - 10:37 am: |
|
Glitch...That bike pipe is full on Deja Vu for me. I had one back in the late 1950's, only it was called a "Popper" at the time and was a slash cut aluminum affair. Made me the envy of all the kids and enemy of all the parents. LOL |
Blake
| Posted on Thursday, April 07, 2005 - 12:51 pm: |
|
Matt, Exactly. And when I have set it to "Never" the darn thing will still sometimes update.
Welcome back EBear. I guess we oughta mention that little logout/login trick somewhere in the error message text. |
Paulinoz
| Posted on Thursday, April 07, 2005 - 07:04 pm: |
|
Is There a mathematical reltionship between. Power, Weight and acceleration. There must be an Injuneer out ther who knows, need to resolve an argument. |
Road_thing
| Posted on Thursday, April 07, 2005 - 08:29 pm: |
|
Absolutely. (More Power) + (Less Weight) = (Faster Acceleration) Anything else I can help you with? rt |
Blake
| Posted on Thursday, April 07, 2005 - 11:55 pm: |
|
Paul, Not with only those parameters. You would also need velocity or some other related parameter to be able to characterize a relationship between power, mass, and acceleration. It takes twice the power to accelerate an object of set mass at the same rate from 30 to 50 mph as from 10 to 30 mph and it takes four times the power to accelerate from 70 to 90 mph at the same rate. In each case velocity has increased by the same delta of 20 mph in the same amount of time (the very definition of constant acceleration). What has changed is the average velocity over which that acceleration has occurred; it went from 20mph to 40 mph to 80 mph, first doubling then quadrupling. Why is that? Well one formulation of power is the measure of a force acting to move an object through a distance over a time interval. When we double the speed, we also double the distance traveled over the same time interval. Thus the power must be double. So the power required to accelerate at a constant rate increases proportionately with velocity, and that is neglecting aerodynamic drag. The power (P) required to accelerate an object at a constant rate (a) at two different velocities, V1 and V2, is P1 for velocity V1, and P2 for velocity V2 where the ratio of P2/P1 is equal to that of V2/V1. Bring significant aerodynamic drag into the equation and the power required to accelerate at a constant rate approaches a proportionality with velocity to the fourth power (V4! Now if you are talking about a better defined scenario like a drag race from a stand-still, then we know the initial velocity; it is zero. We also know the distance; it is a quarter mile or 1320 FT. Ignoring aerodynamic drag we can write a relationship for power and mass wrt the average acceleration for that less ambiguous scenario. Knowing your land speed connection, I'n not sure which you are thinking of, acceleration from a standstill over a set distance, or acceleration at various speeds over some varying distance. Hope you are confused. I tried my worst just for you. (Message edited by blake on April 07, 2005) |
Blake
| Posted on Friday, April 08, 2005 - 12:35 am: |
|
More? Think of power (P) as rear wheel thrusting force (F) times velocity (V) or P=FV. Where the rear wheel thrusting force would be equal to engine torque times the overall drivetrain reduction ratio divided by the height of the rear axle from the road surface, neglecting slippage and friction losses of course. So P=FV So we again see that as velocity increases, the power required to generate the same amount of thrusting force increases proportionately? Hmmm.... We also know that acceleration (a) equals thrusting force (F) divided by mass (m) or... a=F/m So if F is held constant, and m is held constant, then the acceleration (a) also remains constant. But for F to remain constant requires that power increase proportionally with velocity. That's as close to an actual relationship between power, mass, and acceleration as I can get without having to resort to higher math. Of course no motorcycle or car accelerates at a constant rate through the gears. And aerodynamic drag is not insignificant. So pretty much... never mind, it ain't reall except in outer space and then we don't deal in power so much as thrust and all that relativity stuff happens. It's all relative. Cause see if you are traveling due west at a ground speed of 160 mph around noontime, your speed relative to the sun is probably somewhere around 1,200 mph. Also consider that a satellitetraveling due east at 5,822 mph in a circular equatorial orbit around the Earth's axis will have a relative ground-speed of... zero. So you can truthfully say that you went infinitely faster on your Buell compared to a satellite orbiting the Earth. (Message edited by blake on April 08, 2005) |
Paulinoz
| Posted on Friday, April 08, 2005 - 12:49 am: |
|
OK so how many Swallows flying in Tandem can carry the coconut then. What I want to know If I have a M/C with a combined Rider machine weight of 250 Kg and a rear wheel HP of 100.'How much weight would you need to shave of the combined unit to still keep up with another M/C that has 10% more HP and weighs 250 Kg. Or am I being to simplistic which is very much in the cards. By the way who swallowed the coconut. |
Court
| Posted on Friday, April 08, 2005 - 04:31 am: |
|
>>>Ignoring aerodynamic drag But it the REAL world it DOES matter what you wear. |
Blake
| Posted on Friday, April 08, 2005 - 05:11 am: |
|
In a race to peak speed, assuming optimum gearing in all cases, there's nothing you could do assuming the aerodynamic drag is identical between the two bikes/riders. The higher HP bike will have a higher terminal/peak velocity. No matter how heavy, given enough time and distance it would catch and pass your lighter slower machine/rider. If you are talking about a 1/4 mile drag race from a stand still, then the rule of thumb is something like ~7 LBs (~3 kilos) or so dropped is worth 1 HP. So to gain equivalence with the bike that is ten HP up on yours, you'd need to drop round about 70 LBs. Remember the a=F/m from above? You want to maintain the same "a" with an "F" that is 10% less so you need to reduce your mass "m" by 10% to compensate. Checking the rule of thumb, a typical sport bike and rider will weigh around 700 LBs, ten percent of that is 70 LBs. Damned rule of thumb is pretty darn good. Must have been derived from that Jerry Haughton thumb. |
Ravensmith22
| Posted on Friday, April 08, 2005 - 06:27 am: |
|
African, or European? |
Glitch
| Posted on Friday, April 08, 2005 - 09:15 am: |
|
Made me the envy of all the kids and enemy of all the parents. I guess history has repeated itself! His grin is almost as big as mine when he comes in the driveway with his friends following him. He told me yesterday the neighbor lady told him to turn it off. He asked me if he had to, said he didn't want to because it made him faster! I see a lot of me in him, scares the hell out of Ann. |
Wyckedflesh
| Posted on Friday, April 08, 2005 - 10:08 am: |
|
Paul, it has always been my understanding that for every 7lbs lost in rider/bike weight its equal to adding 1hp to the engine |
Firemanjim
| Posted on Friday, April 08, 2005 - 01:48 pm: |
|
Well I wouldn't mind shedding about 2 HP. |
Blake
| Posted on Friday, April 08, 2005 - 02:09 pm: |
|
Wycked, Ed Zachry. With clarification that said rule of thumb applies for drag racing where acceleration is from a stand still and aerodynamic drag is typically secondary to mass in its limiting effect on performance. You know about Ed Zachry don't you? My neighbor lady told me about him. She is apparently afflicted with a disease that was named after him. Interesting story, especially to hear her tell it. Wanna here it? Here it is... She went to a local psychiatrist, Dr. Lee, asking if he could help alleviate her depression which she had apparently been battling most of her life. In diagnosing her, the Dr. had her disrobe then observed her closely while she crawled on hands and knees first forwards then backwards across the floor of the examination room then repeating that exercise three more times. Dr. Lee then told her "Okay you finish; put on crothes prease. I see probrem. No doubt about it; you have crear case of Ed Zachry disease. So sorry but nothing I can do to hep you." Horrified my neighbor asked "What does it mean?" Dr. Lee explained "It mean your face rook Ed Zachry rike your ass." (Message edited by blake on April 08, 2005) |
Rick_a
| Posted on Friday, April 08, 2005 - 02:38 pm: |
|
Can anyone find me a pic of the Buell display with the leaned over XB's? I know I've seen a couple pics of it here...I just can't find 'em. We're trying to get my dealership to buy one and I need proof of its trickness. |
Wyckedflesh
| Posted on Friday, April 08, 2005 - 05:22 pm: |
|
Blake LOL that was hilarious...I can't remember the first time I had heard the 7lb/1hp rule of thumb but it was back when I was a very young kid. Might have been when my Aunt was racing her RS Camero back in the Can Am days and they were argueing about who should drive her brother or her husband based on which one was lighter since they were equally talented. |
Aesquire
| Posted on Friday, April 08, 2005 - 10:44 pm: |
|
I want to lose about 5 hp. That's how many miles on a Airdyne? |
Rick_a
| Posted on Saturday, April 09, 2005 - 09:15 am: |
|
Nevermind. I found one. |
Court
| Posted on Saturday, April 09, 2005 - 09:34 am: |
|
Rick: I've got plenty of you need more just drop me a note, won't waste bandwidth. |
Court
| Posted on Saturday, April 09, 2005 - 09:45 am: |
|
You can find darn near anything on the internet, eh?
|
Phillyblast
| Posted on Saturday, April 09, 2005 - 10:04 am: |
|
Court, I recognize the guy in the middle, but who are the other two folks? Man what a shame about Britten - I can't even imagine what he'd have dreamed up (and produced - and raced - "real genius' ship" ) today. Stunning designs. D |
Court
| Posted on Saturday, April 09, 2005 - 11:41 am: |
|
John Britten was a gift we were robbed of. He and Erik Buell had some of the most interesting conversations. I just got info about some of the "big picture" stuff, but they both had a unique appreciation for "simple REAL engineering", i.e. the thing should, first and foremost, work well. I kinda recall Erik mentioning something that made me think the two of them shared more than one giggle over the number of folks who eschewed simple engineering principles in favor of flashy graphics and suspension designed more by marketing than engineers. Erik and John were to have met at Daytona and tragically John was taken from us shortly before they were to have met. Vickie and I had dinner with Kirsteen about 6 years ago. I never met John Britten and I'm very sorry I didn't. Motorcycling would be different today if the glassblower turned motorcycle designer were still here. Kirsteen (a former international model) on the left Jessica, Isabelle and Sam Britten |
Court
| Posted on Saturday, April 09, 2005 - 11:45 am: |
|
Hey...your comment about the "guy in the middle" reminds me......you gotta hear some of my tales about "Guards I have BS'd into letting Buells into places they have no business being". Three times that come to mind, I have pranced Buells past "NO ENTRY" signs and armed guards. I live for this stuff! Yeah....remind me to tell ya! |
Buellkowski
| Posted on Saturday, April 09, 2005 - 12:44 pm: |
|
Blake, save your asian crap (here and on the pics thread) for your "raid-neyck" bars. It's not becoming of your esteemed reputation as a learned, tolerant gentleman. |
Blake
| Posted on Saturday, April 09, 2005 - 05:08 pm: |
|
Right. I'll shift gears and try to recall some of the old Dago and Polack jokes from my yute growin' up in a Pennsylvania steel town. Yunes'll appreciate them more I suppose. |
Wyckedflesh
| Posted on Saturday, April 09, 2005 - 05:54 pm: |
|
Just imagine what we would be riding if Erik and John had formed a company together... Folks I have to say, Reg put on a great Battletrax event today at Scottsdale BMW. Quite a number of the AZBuellers crowd showed up and for a bit, it was an all Buell event at a BMW dealership...kinda makes me sad that the Buell dealers here just can't seem to get it...but my hats off to Scottsdale BMW and Reg I know my wallet is upset, but I see some future gear purchases from that dealership...and possibly a bike down the road... |
Pilot
| Posted on Saturday, April 09, 2005 - 08:06 pm: |
|
G'Day, I am looking for somebody in the Chicago area to help out a fellow Aussie.He is travelling to Oshkosh and is arriving at O'Hare.He needs information on how he should undertake this part of the journey he will also need to get back to O'Hare so as to continue his travels after the airshow is over.He is staying at the university of Wi.Any help would be greatly appreciated.Thanks Ross |
Court
| Posted on Saturday, April 09, 2005 - 08:44 pm: |
|
Ross: Post details of dates, transport needed from (exp: O'Hare) to (exp: Oshkosh) and so forth. Surely we can conjure up someone to help out. I still recall the day I took two wandering Aussies on an impromptu tour of the Buell factory. All the best to you and Fiona. If your Mate is coming through NYC on the way, I'm available to handle any details you need. Court |
|