Vlad,
You wrote "actual distance traveled" and unless that measuring device is attached to your board, it won`t be *actual.
Better yet, have one attached at both ends of the deck and see if they match?
Now run the course consecutive times and see if the distances match?
The only one true constant is change....
Help with a slalom-related MATH PROBLEM
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Vlad Popov
- Moscow-Washington
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Wesley Tucker
- 1961-2013 (RIP)
- Posts: 3279
- Joined: Tue Aug 27, 2002 2:00 am
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Vlad Popov
- Moscow-Washington
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Vlad Popov
- Moscow-Washington
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Wesley Tucker
- 1961-2013 (RIP)
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Ah, Chuck, it took me a while, but I got it:
3600 = 60 x 60 (Total seconds in an hour)
So 3600 x distance divided by 5280 divided by the time.
Let's see a course is 40 cones at 7 foot spacing with a start and a finish, so let's say the course is 320 feet long?
3600 x 320 = 1,152,000
And let's say somebody lays down a time of 13.28?
5280 x 13.28 = 70118.4
So, 1,152,000/70118.4 = 16.42 MPH
Sounds about right to me.
3600 = 60 x 60 (Total seconds in an hour)
So 3600 x distance divided by 5280 divided by the time.
Let's see a course is 40 cones at 7 foot spacing with a start and a finish, so let's say the course is 320 feet long?
3600 x 320 = 1,152,000
And let's say somebody lays down a time of 13.28?
5280 x 13.28 = 70118.4
So, 1,152,000/70118.4 = 16.42 MPH
Sounds about right to me.
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Pat Chewning
- Pat C.
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This is the way I do it:
60 MPH = 88 FPS
FPS is feet per second.
S = D/T * (60/88)
Where D is distance in feet, T is time in seconds, S is speed in MPH.
This is the same as the previous equation, but I have the 60MPH = 88FPS conversion factor embedded in my memory from using this type of calculation in my engineering work.
This (plus some more math) led me to my prior conclusion that time differences of 1/1000 of a second in a slalom course is the equivalent of about 1/3 of an inch.
MATH: 20 * 1/1000 * (88/60) * 12
Hint: 12 inches per foot
Hint: 20 miles per hour
Extrapolating further:
1/1000 s = .3 inch (approx bearing spacer length)
1/100 s = 3 inches (approx wheel diameter)
1/10 s = 30 inches (approx skateboard length)
1 s = 300 inches (width of neighborhood road)
-- Pat Chewning
<font size=-1>[ This Message was edited by: Pat Chewning on 2003-10-09 02:43 ]</font>
60 MPH = 88 FPS
FPS is feet per second.
S = D/T * (60/88)
Where D is distance in feet, T is time in seconds, S is speed in MPH.
This is the same as the previous equation, but I have the 60MPH = 88FPS conversion factor embedded in my memory from using this type of calculation in my engineering work.
This (plus some more math) led me to my prior conclusion that time differences of 1/1000 of a second in a slalom course is the equivalent of about 1/3 of an inch.
MATH: 20 * 1/1000 * (88/60) * 12
Hint: 12 inches per foot
Hint: 20 miles per hour
Extrapolating further:
1/1000 s = .3 inch (approx bearing spacer length)
1/100 s = 3 inches (approx wheel diameter)
1/10 s = 30 inches (approx skateboard length)
1 s = 300 inches (width of neighborhood road)
-- Pat Chewning
<font size=-1>[ This Message was edited by: Pat Chewning on 2003-10-09 02:43 ]</font>
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Chuck Gill
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- Location: Huntsville, Alabama, USA
Damn liberal arts majors.
Anyway, I have no idea what that equation you posted was all about, but an easier version would be
S = 3600 D / 5280 T
or, simplified
S = 45 D / 66 T
where:
S = speed in miles per hour
D = course length in feet
T = elapsed time in seconds
Note: This will give your apparent forward speed ONLY, not your actual velocity, since the distance travelled in a slalom course will be longer than the point-A-to-point-B length of the course.
Anyway, I have no idea what that equation you posted was all about, but an easier version would be
S = 3600 D / 5280 T
or, simplified
S = 45 D / 66 T
where:
S = speed in miles per hour
D = course length in feet
T = elapsed time in seconds
Note: This will give your apparent forward speed ONLY, not your actual velocity, since the distance travelled in a slalom course will be longer than the point-A-to-point-B length of the course.
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Wesley Tucker
- 1961-2013 (RIP)
- Posts: 3279
- Joined: Tue Aug 27, 2002 2:00 am
Could I trouble someone to help me with a math problem? It's actually an algebra problem, but the difficulty is the same. I majored in Journalism for a reason: no advanced math needed for a degree!
Here's the formula:
60/([5280/D]*[T]/[60])
This, believe it or not, will provide you with an accurate speed in MILES PER HOUR through a slalom course.
D=Actual distance of the course in feet
T=Elapsed time through the course
It translates into dividing the number of feet in a mile by the number of feet in the course, multiplying that number by the elapsed time and then dividing the result by 60. That result is then divided by sixty and the result is Miles Per Hour. Believe me, it works.
My question is if all the brackets, parentheses and divisors are in the right order? It pesters me that you have run a calculation and then save the result then divide 60 by that number. I know there has to be a more fluid, sequential and "algebraic" way of expressing this formula. Unfortunately, I did about as well in Fortran programming as i did in Algebra II, so my formatting skills are minimal at best.
Would someone with a sharper mathematical mind than mine please look at this and edit it so that the result is the same, but the formula is clearer? I originally wrote this down last Spring for Jack Smith who was wanting to calculate actual speeds for the first Paso Robles race. I know it works, but I just have a lingering doubt it can be improved.
Or maybe I got it right the first time and this is the best representation of the problem?
Any help would be appreciated.
Here's the formula:
60/([5280/D]*[T]/[60])
This, believe it or not, will provide you with an accurate speed in MILES PER HOUR through a slalom course.
D=Actual distance of the course in feet
T=Elapsed time through the course
It translates into dividing the number of feet in a mile by the number of feet in the course, multiplying that number by the elapsed time and then dividing the result by 60. That result is then divided by sixty and the result is Miles Per Hour. Believe me, it works.
My question is if all the brackets, parentheses and divisors are in the right order? It pesters me that you have run a calculation and then save the result then divide 60 by that number. I know there has to be a more fluid, sequential and "algebraic" way of expressing this formula. Unfortunately, I did about as well in Fortran programming as i did in Algebra II, so my formatting skills are minimal at best.
Would someone with a sharper mathematical mind than mine please look at this and edit it so that the result is the same, but the formula is clearer? I originally wrote this down last Spring for Jack Smith who was wanting to calculate actual speeds for the first Paso Robles race. I know it works, but I just have a lingering doubt it can be improved.
Or maybe I got it right the first time and this is the best representation of the problem?
Any help would be appreciated.