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Post by SpeedGeek on Jan 21, 2006 13:40:27 GMT -6
Let's start a new thread to bring some thought-provoking questions about the science of woodcar racing. Periodically, we will post a question that requires those interested to (usually) calculate the solution to a real race situation. The intent here is get newbies up to speed faster (pun intended...) and everyone in general to look at the race from a different angle. Depending on its success, we will start out simple and get more complex. There are no prizes, no points awarded; it's just for fun. Here goes:
In last month's race, Red Frog crossed the finish line at an estimated velocity of 11.5 feet per second. At this velocity, what is the rotational speed of the wheels in units of revolutions per minute (RPM)?
To put the answer in perspective, equate this speed to an everday object that rotates at a similar speed.
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Post by WarpSpeedINC on Jan 21, 2006 16:17:34 GMT -6
about 2161 rpm. Assuming his wheels are the wirl maximum diameter of 31mm/1.223".
I think it is pretty close, but I am no mathematician.
Warp Speed
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Post by slkrnsntracing on Jan 21, 2006 18:03:38 GMT -6
Is the actual distance of the race 42'  Phil |
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Post by woodenwonder on Jan 21, 2006 19:10:46 GMT -6
I got 2156 which is close to Warps figures. Probably depends on how far you carry out PI. I used standard figure of 3.14.
Phil,
Length of track does not matter. Only velocity of car and circumference of wheel.
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Post by slkrnsntracing on Jan 21, 2006 19:21:20 GMT -6
Phil, Length of track does not matter. Only velocity of car and circumference of wheel. I understand that, but I still would like to know the distance of the race. Phil |
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Post by wheeler on Jan 21, 2006 19:35:39 GMT -6
PHIL, I believe the dimension from contact side of start pin to center finish line hole is 400.125. They are now using largest finish line hole so I think you add 3 in. to this dimension.
Fred
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Post by slkrnsntracing on Jan 21, 2006 20:12:18 GMT -6
Thanks Fred,
Good Luck tonight.
Phil
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Post by patmanruls on Jan 21, 2006 22:56:33 GMT -6
I say this lovingly... you guys are geeks.....
LOL....... pat... to me.. I'm just singing "the wheels on the car go round and round, round and round, round and round; the wheels on the car go round and round, right down the track........
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Post by patmanruls on Jan 21, 2006 22:56:57 GMT -6
I failed that part of chemistry anyway....
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Post by warthog on Jan 21, 2006 23:00:32 GMT -6
I'm just singing "the wheels on the car go round and round, round and round, round and round; the wheels on the car go round and round, right down the track........ and I thought it was just me.... now if we start talking driver psychology count me in. |
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Post by SpeedGeek on Jan 24, 2006 18:50:39 GMT -6
The horse isn't dead yet... relate that speed (~2200 RPM) to an object that is near and dear to all of us.
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Post by Sssnake on Jan 24, 2006 20:48:30 GMT -6
This is a little divergence from the rpm. The average car tire (15") is 25.4 inches in diameter, the average PWD wheel is 1.18 inches in diameter. This gives you a scale of 1:21.5 . Multiply the length of 40' times your scale of 21.5 = 860' for the track length. Now you have the length and the time, lets use 2.97 sec. Divide 860' by 2.97 seconds = 289.56 ft/sec.
Now the easiest way to convert to MPH, without getting into the whys and wherefores, is simply to divide feet per second by feet in a mile; then times that by 3600.
289.56 / 5280 x 3600 = 197.43 MPH
And if you divide 1320' (quarter mile) scale feet by 289.56 ft/sec. you get 4.56 sec. in the quarter mile.
Not bad for a little wooden toy.
Sssnake
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Post by patmanruls on Jan 24, 2006 21:07:32 GMT -6
Sssnake,
Welcome back!!! I made a post sometime back asking if anybody had heard from you, or what had happened. Last thing I remember, you had some bad luck with the shipping of your cars..... and boom! no more Sssnake....
I cannot remember the name of the car, but it was shiny, and had the rail protectors on it..
Again, welcome back, and I hope to see one of your cars soon, if it didn't make it in January.
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Post by Sssnake on Jan 25, 2006 6:27:34 GMT -6
That was "roller Derby"  been busy with family and business affairs, but there is a "Roller Derby II" Who knows it might make it to WIRL. |
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Post by woodenwonder on Jan 25, 2006 8:04:21 GMT -6
Sssnake,
Welcome back! I would have loved to see how that car would have ran. To be quite honest I am in the process of making one similar. Well with the same concept but it won't look as good as yours. Hope you don't mind that I borrowed your idea.
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Post by parrish on Jan 25, 2006 8:08:41 GMT -6
Welcome back Sssake!
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Post by wheeler on Jan 25, 2006 17:26:19 GMT -6
Clever idea but I think it violates the 3/8 bottom clearance rule. Anything that touches the track other than the wheels could scuff guide rails. Would hate to see new track damaged.
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Post by Smokinjoesracing on Jan 25, 2006 18:14:08 GMT -6
Ditto Fred.
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Post by Sssnake on Jan 25, 2006 19:48:10 GMT -6
Actually it doesn't violate any rules. It has the required 1 3/4" x 3/8" clearance down the middle of the car. The rollers do not touch the track and do not scuff the rail as they rotate.
P.s. The car was extremely fast. I ran it in the outlaw class and the only car that could touch it was a motor driven propeller car that beat it by about one car length.
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Post by slkrnsntracing on Jan 25, 2006 19:52:32 GMT -6
Fred & Smokin',
I don't necessarily care either way whether it's legal or not, but you have to admit SS was thinking.
I don't think the 3/8' clearance rules apply outside the 1 3/4" center rail width. The only thing it says about anything touching the track is that the wheel can't be painted or metal.
In my mind I had taken it for granted that only the wheels could touch the track, but I don't actually see it in the rules.
If the part that touches the center rail could spin, it shouldn't hurt the track. It might even be considered a wheel. The rules only say that a car has to have at least 4 wheels.
Like I said I just liked the thought process.
Phil
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