plz help me

watch this and you can even locate it with 1

well i cant watch videos on youtube could you send it to my email? @Eimerplayz127 xpianozhangx@163.com thank u

1. Triangulation 2.Ninjabrain bot
2. I'm noob and I just want to help

ninjabot

search axis or perfect travel on yt

Well, it depends, there are many methods, for example:

1. Standard triangulaton Simply throw one eye, follow its direction, go a bit to the side and throw another eye. Stronghold is at the intersection of lines created by eyes. Heres a simple tutorial by Couriway:

2. Axis calculated Instead of just gauging stronghold's direction you can work out your distance to stronghold using angle diffrence, coordinates offset and a bit of math. The best way to do it is using Axis Calculated developed by Ninjabrain:

3. Calculated travel Neat thing about throwing eyes is that they will always point towads 8,8 coordinates within a chunk. So, by using special program and very precise angle measurement you can find exactly wich chunk it is pointing towards. Here is a great tutorial by Four:

If you do not like using calculators, then locating the stronghold with just 2 eyes will be very difficult. Without calculators, expect to be using around 5 eyes or so every time. If you don't mind using calculators, then thats fine, the above videos will be perfect for you.

well so what is the expression plz im not good at math @energy980

@CommunismNeverDies what do you mean with "the expression"

also, just make the maths mentaly smh distance, so pitagoras, triangles wont be hard :Keepo:

@KilleDragon i mean that i dont know how to express the x in the function how can i express the angle of the eye like i threw one at x1260z0 and the angle is -147 how can i express it and how can i draw down the function image plz is that x=-1.47y+1260 ?

Throw an eye, note the angle of the eye. Turn 90* and move 17.5 blocks (around 4.5 sprint jumps or so). Throw the eye again and note the second angle. Calculate the difference and then take 1000/(difference of angles). That will give you the distance and direction of the stronghold from your current location. I basically recited what is from here: https://imgur.com/gallery/PUsG4Cc

well i just want to calculate it with function i think its easier to understand @energy980 and i think its not very easy to reach there in the nether im not good at mathematics (or maybe yes?

ninjabrainbot probably the best: https://github.com/Ninjabrain1/Ninjabrain-Bot

YA REALLY WANT TO KNOW HOW TO CALCULATE WITH FUNCTION if ur not good at math then you might not

x1 = x coord 1 z1 = z coord 1 x2 = x coord 2 z2 = z coord 2 a1 = angle 1 a2 = angle 2

lines: 1. y = cotangent(a1)(x - x1) + z1 2. y = cotangent(a2)(x - x2) + z2

the only real way to do this is a calculator or graphing calculator, so you should use ninjabot anyway if you're going to use a calculator

if you really want to use a function, you can use mine: https://www.desmos.com/calculator/hqqabjstkc

full equations (copy paste into desmos scientific or graphing calculator): y=\left{\operatorname{sign}\left(\theta_{1}\right)>0:\cot\left(\theta_{1}\right)\left(x-x_{1}\right)+z_{1}\left{x>x_{1}\right}\left{x^{2}+y^{2}\le24320^{2}\right},\ \cot\left(\theta_{1}\right)\left(x-x_{1}\right)+z_{1}\left{x<x_{1}\right}\left{x^{2}+y^{2}\le24320^{2}\right}\right}

y=\left{\operatorname{sign}\left(\theta_{2}\right)>0:\cot\left(\theta_{2}\right)\left(x-x_{2}\right)+z_{2}\left{x>x_{2}\right}\left{x^{2}+y^{2}\le24320^{2}\right},\ \cot\left(\theta_{2}\right)\left(x-x_{2}\right)+z_{2}\left{x<x_{2}\right}\left{x^{2}+y^{2}\le24320^{2}\right}\right}

Ayo that's a nice email you got

wait a minute can we use calculator when speedrunning?

Yeah man, new rule

@n00bie so yes or no please give me an exact answer

yes you can use

literally just use ninjabrain, f3 + c and bam