Everything posted by ohhungry
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Making a dbow rusher
If dbowers only have 1 pray then screw it, I'll just kill the rushers :p makes more sense to have the pray levels instead of the health though...you can still get 1 shotted and lose your dbow if you don't have 25 pray, but with prot item you can't lose it and my low health makes up for the cb levels anyways?
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Banned twice in 3 weeks.
You underestimate the ability of jagex to detect bots. 3 weeks between bans is pretty decent
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Making a dbow rusher
So I'm making a dbow rusher ATM, got 60 range, 19hp, and 25 pray. I decided to try it out tonight without much success. I'm 37 cb, and going after the obby maulers, but they all have like 50+ hp and I can't seem to spec more than 18. I'm using range pots and red dhide. Is there something I'm missing? Or do I just need to raise my range level more?
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0 view count 0 replies 3 days?
I had the same issue with a thread that I made, then all of a sudden it spiked up to like 80 views, seems like a delay of some sort?
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Selling chocolate dust cheap
I've got about 8.7k, I'm looking for somebody to buy all, price is 160 ea, pm me if interested
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Dynamic IP vs VPN?
It depends on your ISP how long you will have a certain IP, resetting your router won't work with all of them. If you go to your router settings you there is a section that tells you your lease time, which is how long until you get assigned a new IP I think. Mine is 1 day...
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OSBot Gilded Altar
Thank you for making this open source. It's great to have resources like this for learning coding techniques
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Realistic facts about the possibility of me getting unbanned
What? They actually quashed one of your offences? Looks like there might be hope for me too, since mine was on a different ip as well
- [[Hardest Game]]
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Training equipment for my character?
The fastest way you could level your strength would be to get full dharocks, lower your health with rock cakes, then go to mm tunnels or bandits and pray melee. Also wear a fury+rune boots+god cloak. And if you don't wanna do that, get a salve ammy, wear some barrows armour, obby shield, drag scimmy, and kill ankous
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Gf Jagex
I use a VPN, so I technically I am changing ip's? I'm not sure what you mean by same sort of ip, or what you thought I meant by changing ip's in the first place. Jagex should be able to detect VPN use I think, but don't see how they could IP ban me for that
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Realistic facts about the possibility of me getting unbanned
Realistically, they won't read the appeals. There is a small chance they will read the emails, but I wouldn't expect a response for atleast a week or two. I know how you feel though, I'm going through the same process right now, my main got banned like 2 days ago botting pest control. Although the fact that I got like 7 other accounts banned for goldfarming within a couple days of eachother besides that, my circumstances might be a little different :P I was using a vpn on my main the entire time, so I just went on my home ip which my account was made on and said I was on vacation the whole time and must have gotten hacked, but honestly I doubt there is any chance of being unbanned. You might have a chance with that guy who's apparently a jagex mod or something, although I doubt thats legit, its sadly most likely your best chance out of all of those
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Math Challenge # 2 - 100k winner
We can do this with a simple triple integral and some algebra. If you split the cylinder into very small circular slices, each of those slices would have a volume equal to (Area of Circle)*(Δh). To find the area of each circular slice, we can take the circle to be a series of concentric disks, each with a width Δr. Using polar coordinates, we can express the area of the circle as a double integral. Since each disk goes about a 360 degree rotation (or 2π) this will be our angle of integration, giving us the following integral: ∫(0-2π)∫(0-R)[rdrdθ] We now take this integral, and integrate it over our height of 10cm, and this integral will be equal to the volume of the cylinder, which is 300cm^3 So our equation is: 300 = ∫(0-10) ∫(0-2π)∫(0-R)[rdrdθdh] 300 =10 (∫(0-2π)∫(0-R)[rdrdθ]) 30 = 2π(∫(0-R)[rdr]) 30 = 2π[(r^2)/2](0-R) 30=π(r^2) sqrt(30/π) = r r ~ 3.1 Therefore the answer is A) 3.1cm
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Gf Jagex
It wasn't an ip flag, from what I've heard it might have been cause I was changing IP's too often, also I guess there was a bot nuke or something. I've gotten 8 accounts banned now, all for goldfarming. 3 fletchers, 4 fishers, and my main, who was botting pest control at the time(how the hell is that gold farming...) Anyways, I got 2 accounts left, one with decent enough combat stats, and about 6 mill in my pocket, so time to rebuild :P
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Goldfarming bans
What were you botting? An autotyper?Also, I just can't understand how the ip's i've been switching too could be flagged, I'm using HMA and there's like 32000+ ip's. It might be having too many bots running at the same time, so all these accounts on the same ip, but I'm not sure cause I've seen people running like 8 bots at a time and stuff, max I've ever had at once is 5 No I bot smart, or maybe super lucky Well you probably only bot on a main,not quite the same risk :p
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Why do scripters...
most anti-bans I've seen don't change the camera pitch, they just rotate it around, so the pitch shouldnt change. I think the way the methods in the API works though when it tries to rotate the camera to make something visible is it changes the pitch, which is why you see the camera in such weird positions all the time. I tried making a script that would move the pitch back up to the top, but it looks even worse because it keeps moving the pitch down then up over and over and looks retarded...
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Riddle for 1 GP!
well that was over fast, gratz on your 1 gp
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Math Challenge # 1 - 100k winner
the answer i have is based on a school textbook, maybe he is trying to figure out something wrong, or has made a mistake, the answer is 4x + 3 so someone try to explain why. Maybe we get a prize for telling you how you are wrong? Obviously your textbook is wrong or you're just trolling everybody... if 4x+3 was a factor, and we put it in the form (x-s)(x-t) you would then have (4x+3)(x-t)=4x^2-8x+3 Now if you expand out those brackets, the only way to have 3 on both sides is if t= -1, and if t= -1, there will -x on the left side and -8x on the right side, and the equation would obviously not be true for all values of x, hence 4x+3 is not a factor.
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Math Challenge # 1 - 100k winner
There were two plausable choices with that equation, I know I figured it out faster then you did Nah, I knew the answer already, but I wanted the longest description cause apparently thats the contest or something :P
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Riddle for 1 GP!
First correct answer wins an astounding 1 gold coin. A man is doing a science experiment in the Arctic. His experiment involves determining how much power it takes for different drills to dig a hole through the ice to a depth of 5 metres. He has 3 different drills to test. The first one is 5 meters in length, with a radius of 2 metres at the base, and makes an angle of 40 degrees from the centre of the tip to the base of the drill. The second one is similar, but instead makes an angle of 30 degrees, and the third an angle of 20degrees. Each one uses electrical power (created by a generator and solar panels) based on the ratio of the angle (in radians) to the radius of the drill. The scientist must decide which drill uses the least power, but before he starts, he sees a bear walking towards him. What colour is the bear? Explain your answer
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Math Challenge # 1 - 100k winner
ahha nope sorry try again, note: ^ means exponent He's actually right... and he did it in a manner much harder then it needed to be. Pfft, trial and error is for nubs
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Math Challenge # 1 - 100k winner
ahha nope sorry try again, note: ^ means exponent Uhh...yeah, my answer's right. unless by 4x^2 you actually meant (4x)^2, in which case there are no real roots.
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Math Challenge # 1 - 100k winner
4x^2-8x+3 to find roots, use quadratic formula x=(8(+/-)sqrt(8^2-4(4)(3))/8) =(8(+/-)sqrt(16))/8 =(8(+/-)4)/8 =3/2 and 1/2 Set all options equal to 0 and sub in either value of x, whichever statements hold true are correct a)2x+1= 0 (does not work for either value of x) b)4x+3 = 0 (does not work for either value of x) c)2x-3 = 0 (work for x=3/2, therefore this is a factor of 4x^2-8x+3 d)2x+3 = 0 (does not work for either value of x) Therefore 2x-3 is a factor of 4x^2-8x+3, the answer is C Sidenote: You should probably choose harder math problems :P
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MoneyMan's Money Service
You're just trying to buy feedback. If this was allowed feedback wouldn't mean shit.
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Goldfarming bans
Well, I sent an email to my vpn provider to see if they could provide me with a static IP, I'll see how it goes