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10% on a treadmill for 3 miles... vertical is?
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Ok, can't remember how to the do the whole sin/cosin thing or even if it is much more simple than that. Perhaps (5280x3)x.10? Little help here.

10% incline on a treadmill for 3 miles is how many vertical feet?


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Last edited by: bloomers: Jan 6, 11 15:39
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Re: 10% on a treadmill for 3 miles... vertical is? [bloomers] [ In reply to ]
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Sin x degrees = vertical distance / total distance

so Vertical Distance = Total Distance * Sin (x degrees)

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Re: 10% on a treadmill for 3 miles... vertical is? [DeepDish] [ In reply to ]
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I allways thought that was 10% grade

jaretj
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Re: 10% on a treadmill for 3 miles... vertical is? [jaretj] [ In reply to ]
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I think so. If that is the case, than it would be .10 of 90 correct, or 9 degrees?


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Yes, I shave my legs. Yes, I am comfortable with this. I am enlightened.
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Re: 10% on a treadmill for 3 miles... vertical is? [bloomers] [ In reply to ]
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I was thinking of a 1 foot rise every 10 feet or .3 miles for 3 miles which would be 1584 feet.
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Re: 10% on a treadmill for 3 miles... vertical is? [DeepDish] [ In reply to ]
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DeepDish wrote:
Sin x degrees = vertical distance / total distance

so Vertical Distance = Total Distance * Sin (x degrees)

That would be for a X degree angle not X% grade.

Hugh

Genetics load the gun, lifestyle pulls the trigger.
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Re: 10% on a treadmill for 3 miles... vertical is? [jaretj] [ In reply to ]
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That's how I computed it in my OP, but I wasn't sure if that was indeed the case.


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Re: 10% on a treadmill for 3 miles... vertical is? [bloomers] [ In reply to ]
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rise/run == % grade = .1
hypotenuse = 3mi.
tan(angle) = rise/run
arctan(rise/run) = angle
sin(angle) = rise/distance traveled
rise = sin(angle) * distance traveled
rise = sin(arctan(rise/run)) * distance traveled
rise = sin(arctan(.1)) * 3mi
rise = sin(5.71) * 3mi
rise = .09949 * 3mi
rise = 1576'

(for moderate grades, the rise is very close to the grade * distance traveled)

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Re: 10% on a treadmill for 3 miles... vertical is? [bloomers] [ In reply to ]
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If I recall correctly percent grade = rise divided by run (horizontal distance). What the treadmill tells you is the hypotenuse of that triangle so...

X = horizontal distance
.1X = vertical distance


(3*5280)^2 = x^2 + .1X^2
(3*5280)^2= 1.01X^2
15761.3=X

Vertical Distance = 1576 feet
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Re: 10% on a treadmill for 3 miles... vertical is? [bloomers] [ In reply to ]
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bloomers wrote:
10% incline on a treadmill for 3 miles is how many vertical feet?

Did you graduate high school? College? If college, what major? Really, I'm curious.
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Re: 10% on a treadmill for 3 miles... vertical is? [BobbyShaftoe] [ In reply to ]
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BobbyShaftoe wrote:
bloomers wrote:
10% incline on a treadmill for 3 miles is how many vertical feet?


Did you graduate high school? College? If college, what major? Really, I'm curious.

I don't know about the OP, but I'll hazard to guess that you didn't take any courses in tact.
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Re: 10% on a treadmill for 3 miles... vertical is? [JollyRogers] [ In reply to ]
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bingo --

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Re: 10% on a treadmill for 3 miles... vertical is? [RhymeAndReason] [ In reply to ]
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RhymeAndReason wrote:
I don't know about the OP, but I'll hazard to guess that you didn't take any courses in tact.

You're right I didn't but my question stands. How can a person have any education at all and not be able to calculate 10% of 3 miles? I know my daughter could do that in the 4th grade.
Bobby
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Re: 10% on a treadmill for 3 miles... vertical is? [BobbyShaftoe] [ In reply to ]
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Dumbass says what? Did you even read the OP's post?


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Re: 10% on a treadmill for 3 miles... vertical is? [bloomers] [ In reply to ]
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A 10% grade is equal to 10 feet of rise for every 100 feet traveled horizontally. This is calculated as follows:

(10 feet) / (100 feet) = 0.10 = 10%
In order to calculate the total vertical distance you traveled, all you have to do is multiply the grade by the distance run. Here's the calculation for your 3 mile run:

(3 miles) * ((5280 feet)/(1 mile)) * 0.10 = 1584 feet

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Re: 10% on a treadmill for 3 miles... vertical is? [Nailey13] [ In reply to ]
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Yup. He first shows what he thinks might be right (and was) but then goes on to say "10% incline on a treadmill for 3 miles is how many vertical feet?" So here is a person who does not have the math skills to be confident in a grade school math problem. All I am asking is: How much education does this person have and in what field? Is that upsetting to you?

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Re: 10% on a treadmill for 3 miles... vertical is? [BobbyShaftoe] [ In reply to ]
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BobbyShaftoe wrote:
RhymeAndReason wrote:
I don't know about the OP, but I'll hazard to guess that you didn't take any courses in tact.


You're right I didn't but my question stands. How can a person have any education at all and not be able to calculate 10% of 3 miles? I know my daughter could do that in the 4th grade.
Bobby

Bobby, whether or not your daughter could figure it out, 10% of 3 miles is not the correct answer.
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Re: 10% on a treadmill for 3 miles... vertical is? [BobbyShaftoe] [ In reply to ]
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Sometimes the answer is not what it seems. Couldn't it be zero? Incline changes the feel of a treadmill, but there's no net vertical involved.
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Re: 10% on a treadmill for 3 miles... vertical is? [pedaller] [ In reply to ]
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pedaller wrote:
Sometimes the answer is not what it seems. Couldn't it be zero? Incline changes the feel of a treadmill, but there's no net vertical involved.

Well, you can take that route, but then there's no net horizontal involved either so...
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Re: 10% on a treadmill for 3 miles... vertical is? [JollyRogers] [ In reply to ]
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JollyRogers wrote:
10% of 3 miles is not the correct answer.

Not exactly, but its within one half of one percent (1576 vs. 1584) and doesn't require the sin of the arcrtangent of the grade. I reckon < half a percent should be good enough for most work-out related issues.
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Re: 10% on a treadmill for 3 miles... vertical is? [JollyRogers] [ In reply to ]
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LOL - most pointless workout ever!
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Re: 10% on a treadmill for 3 miles... vertical is? [BobbyShaftoe] [ In reply to ]
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BobbyShaftoe wrote:
JollyRogers wrote:
10% of 3 miles is not the correct answer.


Not exactly, but its within one half of one percent (1576 vs. 1584) and doesn't require the sin of the arcrtangent of the grade. I reckon < half a percent should be good enough for most work-out related issues.

Trig functions not required, just multiplication, division, squares and square roots - pythagorean theorem. Granted, it's not fourth grade material, but high school for sure.
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Re: 10% on a treadmill for 3 miles... vertical is? [JollyRogers] [ In reply to ]
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Where is rappstar when we need him? He is an engineer, he would know this stuff.
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Re: 10% on a treadmill for 3 miles... vertical is? [BobbyShaftoe] [ In reply to ]
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BobbyShaftoe wrote:
bloomers wrote:
10% incline on a treadmill for 3 miles is how many vertical feet?


Did you graduate high school? College? If college, what major? Really, I'm curious.

Don't know about the op, but I have a doctoral degree and didn't know the answer. And no, I never took many math classes. Cryptonomicon is a great book, but not every " educated" individual has a background in math like most of those characters. I'll hazard that there are a great many areas you have a limited background in that would make you look like a dipshit to those that do.

While I'm sure we all tremble in the presence of your 4th grader's mathematical prowess, it's not exactly like the op asked for the answer to 4+4.
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Re: 10% on a treadmill for 3 miles... vertical is? [BobbyShaftoe] [ In reply to ]
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Fail

Precent grade in the united states is expressed as:

100 x RISE/RUN
Last edited by: LuckyMe: Jan 6, 11 19:59
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