

Why not see how far you can run in 11:11:11 on 111111?

Too long!

Onceinalifetime wrote:
Tomorrow is 11111. Ran a marathon on 101010. Need something special to do. Maybe I'll run as far as I can in 1:11:11. Join me?
I think this is a great idea! I will be joining. The snow might make it tough for me here though. 
This is a GREAT idea. Looking forward to the responses.

Try running 1 mile in 11:11!!

needs to be 11.1 miles, with a 11,111' climb in 1:11:11 on 11111

[quote]lingeron wrote:
Why not see how far you can run in 11:11:11 on 111111?[/quote)
I would do this date for sure. I'm tapering right now, so tomorrow won't work for me. This will be interesting to follow... 
I think that is a great idea too. I will be joining you. However, we have 10 inches of snow since yesterday and it's still snowing so it may be a slow go.

planB. wrote:
needs to be 11.1 miles, with a 11,111' climb in 1:11:11 on 11111
Just calculated the incline you'd have to set your treadmill to to do this and if I've done my math correctly it's an.... 11% incline. 
As much as I want to believe you, my calculations say ~19%...
Kyp Lyttyn wrote:
Just calculated the incline you'd have to set your treadmill to to do this and if I've done my math correctly it's an.... 11% incline. 
This whole idea is stupid. Just run 20 miles. Doubles.

Huh. That sounds like about what I can run a half marathon in right now. However, I'm racing one on Sunday, so that won't be 1/11/11. Too bad!

For 111111 run 11 miles in the morning, afternoon, and evening. The weather should be agreeable for most in the US.

I ran as far as I could in one minute on the first. Made it a little less than 450 meters.

I'll try tomorrow morning to ruin 11.11 miles in 1:11:11 on 11111. Yeah, a 6:24 pace!

duckshirt wrote:
As much as I want to believe you, my calculations say ~19%...
Kyp Lyttyn wrote:
Just calculated the incline you'd have to set your treadmill to to do this and if I've done my math correctly it's an.... 11% incline.
Tell me where I'm going wrong...
using right triangle trig, the hypotenuse of the triangle, representing the running path to 11,111' feet, is 11.1 mi in length.
The vertical side of the triangle is 2.104 mi (11,111') in length.
Solving for the base of the triangle: b = sqrt (11.1^2  2.04^2) = 10.89876983883961.
So now to get the angle between the hypotenuse and base we need the arc tangent of the vertical side / hypotenuse = atan(0.19304931025353339227) = 10.92649759967522612984511395 
I am running from 11:11am to 1:11 pm

You figured out the same thing here:
"vertical side / hypotenuse = 0.19304931025353339227 "
I modified your equation slightly to show you calculated a 19% incline. Keep in mind that percent incline and degrees of an angle are different. Percent incline is distance of rise/distance of horizontal travel. So, a 19% incline has you rise 19 feet for every 100 feet of horizontal travel.
Actual overground distance traveled would then be sqrt(100^2 + 19^2)=101.8 feet. You are running the distance of the hypotenuse if you will. 
The Stache wrote:
You figured out the same thing here:
"vertical side / hypotenuse = 0.19304931025353339227 "
I modified your equation slightly to show you calculated a 19% incline. Keep in mind that percent incline and degrees of an angle are different. Percent incline is distance of rise/distance of horizontal travel. So, a 19% incline has you rise 19 feet for every 100 feet of horizontal travel.
Actual overground distance traveled would then be sqrt(100^2 + 19^2)=101.8 feet. You are running the distance of the hypotenuse if you will.
Well screw me raw. All these years I've been using a treadmill and had assumed that a 5% incline meant that I was climbing up a 5 degree angle. That does make the math much easier to calculate the elevation gain on a treadmill, but I now have to modify my original statment:
Just calculated the degree incline of the grade that you'd have to climb and it's an...11 degreee climb!