[Useless info?] Mount or run?

[Happy2k]

"Is it worth mounting up for getting over there? Or should i just run? What is the fastest?" Well..

I was bored and did the math (soooo hard math...).
Normal running speed is 100% (x), and epic ground mount is 200% (x*2), while it takes 3 sec to mount (-3).

So that leads us to this: x = (x * 2) - 3
For those two be equal to eachother, x would have to be 3.
That concludes:
After running for 3 seconds (after mouting up), you will have gotten as far as the person running on feet. Total time: 6 seconds.

For 280%: After 4,667 seconds
For 310%: After 4,429 seconds

[stormer]

This is equal useless than this camera

[phstyrer]

Epic reply, +Rep for the uberpwnage pic

[asi29]

hehe thx, now i know if i should run from AH to orgri mailbox

[b2zeldafreak]

or just roll a druid

[XinuX]

Lol, I always wonder wich will be fastest :P +rep

[Happy2k]

Lol, I always wonder wich will be fastest :P +rep

Ty ^^ Simple math, but it works

[suran37]

i knew that if you run more then 3 secs its faster to mount that should be common knowledge not something you actually have to sit down and try to figure out

[Qwiz]

Is this really correct?
If you only have to run for 3 seconds to your goal you saved yourself 1,5sec for not mounting and running (3 sec to mount and another 1,5sec to reach the goal). Meaning that you will save time by not mounting when you have to run for <6 seconds, and save time by mount if you have to run for >6 seconds.
As you reach the same distance running without a mount for 6 seconds as if you mount and run(mounted, 3 seconds to mount and 3 seconds to move at 2xrate).

[Jaden2045]

That math actually isnt correct because you arent using the same units while counting seconds and speed.
The correct way to do this would be:

Variables:
Speed - known (s)
Distance traveled where equal - unknown (d)
Timedelay - known (t)

(100(s) * d) - 0(t) = (200(s) * d) - 3(t)

Add 3(t) to both sides
Divide by d
divide by 100s

You are left with:
2(s)*3(t)

Because you are moving a certain speed over that amount of time, the variable (s) can now be removed, leaving 2*3(t)=6(t)
And because we started with t=1second (as by the mounting time), it is more efficent to mount on an epic mount if it will take you more than 6 seconds to reach your target while running.

EDIT:
Just to prove it further, lets use logic. Lets say the 100% speed moves you 100 feet a second, and likewise 200% moves you 200 feet. In the 3 seconds that you mount, the runner moves 300 feet. Then, over the next 3 seconds you are moving at 100 feet faster per second, which would mean it takes exactly 3 more seconds to catch up with him (300 foot lead). 3+3=6.

[Happy2k]

Jaden2045 --> I can see whats wrong with my post.. I didnt quite make clear what ive calculated. The 3 seconds is the time AFTER mouting up. After 3 seconds you will be equal to the person running on feet.

Ill edit post and make that clear


Btw.. You dont need 100s and 200s.. We know that running on feet is 100%, mount is 200%. 1x and 2x - doesnt need to be more complicated than that.

[Jaden2045]

You're right, it could be much simpler
I just wanted to show it in a very layed-out manner, so people could see everything going on. Thanks for clarifying your post.

[Happy2k]

You're right, it could be much simpler
I just wanted to show it in a very layed-out manner, so people could see everything going on. Thanks for clarifying your post.

At least now people can understand my post better Thanks to you too.

[Haitamo]

geek talk -_- i feel like i wanna suicide

[Killa Queen!!1!]

hehe thx, now i know if i should run from AH to orgri mailbox

Well looking at your post i would say mount up since it looks so much cooler than running over