| Author |
Message |
Hans E. Hansen (4re_gt4)
Member
Username: 4re_gt4
Post Number: 716
Registered: 4-2002
| | Posted on Friday, January 03, 2003 - 6:31 pm: | |
Yeah, but what if your RPM increases by 6 inches? Will you be going 2 MPH faster? |
Joel Belser (Driver)
New member
Username: Driver
Post Number: 27
Registered: 9-2002
| | Posted on Friday, January 03, 2003 - 5:41 pm: | |
Well kids...the formula for calculating the speed of your car is: RPM x Wheel Diameter (inches)
____________________________
Gear Ratio x 336
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DES (Sickspeed)
Member
Username: Sickspeed
Post Number: 757
Registered: 8-2002
| | Posted on Friday, January 03, 2003 - 2:55 pm: | |
"co-tangent"
LMAO...! |
Norm Plaistowe (Normp)
Junior Member
Username: Normp
Post Number: 106
Registered: 11-2001
| | Posted on Friday, January 03, 2003 - 2:47 pm: | |
My bad, I thought I was on a co-tangent |
Horsefly (Arlie)
Member
Username: Arlie
Post Number: 553
Registered: 5-2002
| | Posted on Friday, January 03, 2003 - 2:42 pm: | |
Norm, stay on topic; don't go off on a tangent. |
Norm Plaistowe (Normp)
Junior Member
Username: Normp
Post Number: 103
Registered: 11-2001
| | Posted on Friday, January 03, 2003 - 2:29 pm: | |
I'm really confused, I thought Radius and Circumference were just friends. Now I see that they are in a relationship.
(somebody had to do it) |
Fred (I Luv 4REs) (Iluv4res)
Member
Username: Iluv4res
Post Number: 291
Registered: 8-2002
| | Posted on Friday, January 03, 2003 - 1:17 pm: | |
ha,ha, ha, ha, ha.........
I started reading this thread and immediately stopped. Too much brainpower to go through it!!! |
Lawrence Coppari (Lawrence)
Member
Username: Lawrence
Post Number: 447
Registered: 4-2002
| | Posted on Friday, January 03, 2003 - 11:02 am: | |
At the risk of sounding pedantic, take the first derivative of C with respect to R in the formula C = 2*PI*R. The R vanishes meaning the increase in radius is independent of circle size.
dC/DR = 2*PI |
Mitch Alsup (Mitch_alsup)
Member
Username: Mitch_alsup
Post Number: 281
Registered: 4-2002
| | Posted on Friday, January 03, 2003 - 10:10 am: | |
You have not gone wrong.
Now imagine a string streched around the sun (or universe) and you add 2*pi() feet to the length of the string, the string will now sit 1 foot above the sun (universe). The size of the object only has to be round, its actual size is not necessary. |
Tim N (Timn88)
Intermediate Member
Username: Timn88
Post Number: 1848
Registered: 6-2001
| | Posted on Thursday, January 02, 2003 - 9:59 pm: | |
Hans, that is true? |
Bruno (Originalsinner)
Member
Username: Originalsinner
Post Number: 850
Registered: 5-2002
| | Posted on Thursday, January 02, 2003 - 8:49 pm: | |
dats sum fancy cyphering.I dun learnt it in the 6.th grade just before grad-u-ation. |
Horsefly (Arlie)
Member
Username: Arlie
Post Number: 547
Registered: 5-2002
| | Posted on Thursday, January 02, 2003 - 8:05 pm: | |
If I load a 308 Ferrari with blue fin tuna and leave Chicago at noon doing 125MPH, how long will it be before the timing belts break? |
Hans E. Hansen (4re_gt4)
Member
Username: 4re_gt4
Post Number: 712
Registered: 4-2002
| | Posted on Thursday, January 02, 2003 - 7:27 pm: | |
This was an old puzzle from Junior High math days. The riddle was that if you had a railroad track around the earth, and the temp rose a little so that the rails expanded by 6 inches, how high up off the earth would the now-longer rails be (assuming it was supported so that it rose up an even amount all over). The answer is approx 2 inches. |
'75 308 GT4 (Peter)
Intermediate Member
Username: Peter
Post Number: 2344
Registered: 12-2000
| | Posted on Wednesday, January 01, 2003 - 12:13 am: | |
Thanks!
And Happy New Year! |
Steve Magnusson (91tr)
Intermediate Member
Username: 91tr
Post Number: 1277
Registered: 1-2001
| | Posted on Wednesday, January 01, 2003 - 12:10 am: | |
You went wrong with "...� pi � pi" which is divide by pi^2 instead of divide by 2*pi (i.e., should be 126,720,000 � 2 � pi) |
'75 308 GT4 (Peter)
Intermediate Member
Username: Peter
Post Number: 2343
Registered: 12-2000
| | Posted on Tuesday, December 31, 2002 - 11:56 pm: | |
So where did I go wrong? |
Jeffrey Caspar (Jcaspar1)
Junior Member
Username: Jcaspar1
Post Number: 54
Registered: 5-2002
| | Posted on Tuesday, December 31, 2002 - 11:44 pm: | |
Sorry, I read that backwards. I thought you were increasing the radius by10 not the circumference by 10.
You are correct:
C= 2 x pi x R
C'= C + 10 = 2 x pi x R'
C'-C=10 = (2 x pi x R') - (2 x pi x R)
10 = 2 x pi x (R' - R)
10/(2 x pi) = R'-R
R' -R= 5/pi = 1.59
No matter what the size of the cricle, an increase in circumference of 10 feet will increase the radius by 1.6 feet.
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Jeffrey Caspar (Jcaspar1)
Junior Member
Username: Jcaspar1
Post Number: 53
Registered: 5-2002
| | Posted on Tuesday, December 31, 2002 - 11:36 pm: | |
Defintely off Topic: You mixed up your variables a bit but your basic question remains a good one.
Set Circ of earth = C
Circ 10' above earth = C'
R = radius of the earth
C = 2 x pi x R
C' = 2 x pi x (R+10)
C' - C = (2 x pi x(R + 10)) - (2 x pi x R)
= (2 x pi) x (R + 10 -R)
= 2 x pi x 10
.... C'-C= 62.8 feet
Yes, it is apparently true that by increasing the radius of Arco Arena (where the Sacramento Kings play) by 10 feet, it's circumference increases the same as the extra distance of increasing the earth's radius by 10 feet! Very counter-intuitive but it must be correct.
Actually, it makes perfect since if you think about a square, if you increase a square of any size, no matter what the size, by 10 feet you will increase it's circumference by 10' x 4 sides or 40 feet. Doesn't matter if it is a 8x8' or 1000 square miles!
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'75 308 GT4 (Peter)
Intermediate Member
Username: Peter
Post Number: 2340
Registered: 12-2000
| | Posted on Tuesday, December 31, 2002 - 11:32 pm: | |
1.6ft seems alot to me with only an additional 10ft to the circumference of the earth.
Please bear with me as I'm only a welder...
1mi = 5280ft
24000mi = 126,720,000ft
R of 126,720,000 = 126,720,000 ÷ p ÷ p
R of 126,720,000 is 12,839,420.39
R of 126,720,010 is 12,839,421.40
So I get 1.01ft of radius increase
???? |
Henryk (Henryk)
Member
Username: Henryk
Post Number: 375
Registered: 8-2001
| | Posted on Tuesday, December 31, 2002 - 11:22 pm: | |
James: According to my calculations, the difference in radius, between a C of 24,000 and that of 24,010, would only be 0.0022 |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 292
Registered: 6-2002
| | Posted on Tuesday, December 31, 2002 - 9:43 pm: | |
While a bit off-topic, it has some application vis-�-vis speedometer accuracy with tire wear.
Here is the problem, and the question will become self-evident. Warning, while using basic mathematics, those not interested in math or geometry should exit now.
Thought Experiment:
Stretch a string around the earth so that it is just snug. Now add 10 feet to the length of the string. How much did the radius of the circle formed by the string increase?
(Circumference-of-earth) = 2 x pi x Radius-of-Earth
(Circumference-of-earth plus 10 feet) = 2 x pi x Radius10 where Radius10 equals the radius necessary to increase the circumference by 10 feet.
Therefore, (Circumference-of-Earth + 10 feet) � (Circumference-of-Earth) =
(2 x pi x Radius-of-Earth10) � (2 x pi x Radius-of-Earth)
or
10 feet = 2 x pi x (Radius-of-Earth10 � Radius-of-Earth)
Rearranging:
Radius-of-Earth10 � Radius-of-Earth = 10 / (2 x pi)
Difference in Radius = 5 / pi = 1.59155 feet = 1.6 feet
This means that for an increase in circumference from 24,000 miles to 24,000 plus 10 feet, the radius must increase by 1.6 feet!
Now here is the question. No where in the resulting equation (Difference in Radius = 5 / pi) is there an absolute number describing the circumference or radius of the earth. This same equation would describe the necessary radius increase if one were to stretch a string around Yankee Stadium, and then increase the length of string by 10 feet.
Where have I gone wrong?
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