Cap'n Cunt Posted June 9, 2019 Report Share Posted June 9, 2019 Does no-one else see this Gove/cocaine crap as a pathetic attempt to show that he's human, with human failings, but really just a nice bloke\ he's 'down with the kids'\ 'Hey, I'm a fucking drug addict, all you smack heads should vote for me'? Attention-seeking, desperate, cunt-faced cunt. I wouldn't piss on him if he was on fire. Unless I could piss petrol. I want him dead, and soon. 1 Quote Link to comment Share on other sites More sharing options...
Cap'n Cunt Posted June 9, 2019 Report Share Posted June 9, 2019 And he's a smarmy cunt. I forgot to mention that. Quote Link to comment Share on other sites More sharing options...
Guest judgetwi Posted June 9, 2019 Report Share Posted June 9, 2019 4 hours ago, Cap'n Cunt said: Does no-one else see this Gove/cocaine crap as a pathetic attempt to show that he's human, with human failings, but really just a nice bloke\ he's 'down with the kids'\ 'Hey, I'm a fucking drug addict, all you smack heads should vote for me'? Attention-seeking, desperate, cunt-faced cunt. I wouldn't piss on him if he was on fire. Unless I could piss petrol. I want him dead, and soon. To be fair, cocaine use is an acceptable explanation for looking like a cunt who has been floating face down in a polluted river for three fucking weeks. Personally, I don’t give a shit since I don’t believe a single word that comes out of this back stabbing little creep’s fish like mouth. Gove is that cunt at school who said: ”Miss, Miss, Judge says he loves you and wants to marry you.” ”Oh that’s very nice Judge but i’m afraid i’m already engaged.” ”Yeah but Miss, Judge says he wants to do you up the shitter and spunk all over your face and tits. I don’t know what that means but that’s what he said.” It doesn’t matter how many times you slap Gove after school he’ll never change. Just a natural born creepy little cunt. Oh, by the way, I never said that about Miss Imrie. Gove lied. The cunt. Quote Link to comment Share on other sites More sharing options...
and Posted June 9, 2019 Report Share Posted June 9, 2019 19 hours ago, King Billy said: Is Diane Abbot a mule? A hippopotamus would be closer to the truth. 1 Quote Link to comment Share on other sites More sharing options...
Guest Erroreptile404 Posted June 9, 2019 Report Share Posted June 9, 2019 14 hours ago, judgetwi said: It’s nobody else’s fault that you have the attention span of a retarded chimpanzee. May I recommend the children’s section at your local library. Be aware that they will expect you to return the books within 3 weeks so best to take only one book out at a time. I would avoid Harry Potter if I were you. Probably a bit above you concentration level. No need to thank me! I can just imagine you and your new best friend billy the flid's Facebook timelines, scroll after scroll of pro-Trump, Farage, Tommy Robinson bollocks and memes about "lefties" and "snowflakes". More a case of i couldn't be bothered to read an utterly uninteresting shit nomination submitted by an utterly uninteresting thick cunt. Quote Link to comment Share on other sites More sharing options...
King Billy Posted June 9, 2019 Author Report Share Posted June 9, 2019 25 minutes ago, Erroreptile404 said: I can just imagine you and your new best friend billy the flid's Facebook timelines, scroll after scroll of pro-Trump, Farage, Tommy Robinson bollocks and memes about "lefties" and "snowflakes". More a case of i couldn't be bothered to read an utterly uninteresting shit nomination submitted by an utterly uninteresting thick cunt. More predictable shite from one half of the ‘ranting fuckweeds’ Go and have a lie down. It’ll be time for your medication soon, you thick, incontinent, piss soaked cunt. Quote Link to comment Share on other sites More sharing options...
Guest Erroreptile404 Posted June 9, 2019 Report Share Posted June 9, 2019 5 minutes ago, King Billy said: More predictable shite from one half of the ‘ranting fuckweeds’ Go and have a lie down. It’ll be time for your medication soon, you thick, incontinent, piss soaked cunt. Are you going to cry little Billy and run away again like you did last time? Quote Link to comment Share on other sites More sharing options...
King Billy Posted June 9, 2019 Author Report Share Posted June 9, 2019 10 minutes ago, Erroreptile404 said: Are you going to cry little Billy and run away again like you did last time? Find any nice dogshit today? Do you and your twin BTY have any other interests other than collecting and talking crap? Two of the saddest individuals imaginable ☹️☹️ Quote Link to comment Share on other sites More sharing options...
Guest Erroreptile404 Posted June 9, 2019 Report Share Posted June 9, 2019 12 minutes ago, King Billy said: Find any nice dogshit today? Do you and your twin BTY have any other interests other than collecting and talking crap? Two of the saddest individuals imaginable ☹️☹️ Is this you billy? Quote Link to comment Share on other sites More sharing options...
King Billy Posted June 9, 2019 Author Report Share Posted June 9, 2019 7 minutes ago, Erroreptile404 said: Is this you billy? Remember which ID you’re using today fuckwit. You’re using the other ones insults. Dipstick Cunt Quote Link to comment Share on other sites More sharing options...
Guest Earl Albert of Ross (Bt) Posted June 9, 2019 Report Share Posted June 9, 2019 59 minutes ago, King Billy said: More predictable shite from one half of the ‘ranting fuckweeds’ Go and have a lie down. It’ll be time for your medication soon, you thick, incontinent, piss soaked cunt. Are you talking about Eric the Humongous Cunt who lives in a bus shelter? Quote Link to comment Share on other sites More sharing options...
Guest Erroreptile404 Posted June 9, 2019 Report Share Posted June 9, 2019 1 hour ago, King Billy said: Remember which ID you’re using today fuckwit. You’re using the other ones insults. Dipstick Cunt Of course i'm using their insults because i can just picture you as a bowl cut dumb and dumber type fucking idiot. Paranoid mong. Quote Link to comment Share on other sites More sharing options...
Guest Earl Albert of Ross (Bt) Posted June 9, 2019 Report Share Posted June 9, 2019 15 minutes ago, Erroreptile404 said: Of course i'm using their insults because i can just picture you as a bowl cut dumb and dumber type fucking idiot. Paranoid mong. Everyone on here is a cunt. Quote Link to comment Share on other sites More sharing options...
King Billy Posted June 9, 2019 Author Report Share Posted June 9, 2019 48 minutes ago, Erroreptile404 said: Of course i'm using their insults because i can just picture you as a bowl cut dumb and dumber type fucking idiot. Paranoid mong. Have you ever had any original thoughts? Get back to me when someone’s explained the concept to you. Better still don’t you silly little boy. Quote Link to comment Share on other sites More sharing options...
Penny Farthing Posted June 10, 2019 Report Share Posted June 10, 2019 Gove has said that he would replace VAT with a simpler purchase tax. Boris has said that he will cut higher rate tax. Of course both of them would concentrate on reducing tax for the super rich. However VAT must be one of the most incredibly stupid taxes on the planet and replacing it with a simple non-reclaimable low rate sales tax levied on every transaction would make life a whole lot simpler for small businesses. Quote Link to comment Share on other sites More sharing options...
Penny Farthing Posted June 10, 2019 Report Share Posted June 10, 2019 On 08/06/2019 at 19:47, Major Cunt said: What are you up to this evening Pen? Also could you clarify your gender (old school definition i.e male/female), I can't keep up with the current gender trends. Ta How many angels can dance on the head of a pin? First look at your watch or whatever you use to tell the time and note down what time it is. Then and check the name of the site, go and look at all the cunts who post on here and look at their profiles, then go and look at some of their posts and check back on their profiles. You should by then realise that most if not all of them are not who or what they say they are. After doing this check the time and you should come to the realisation that you have just wasted several hours doing something that will not improve your life in any way .. time that you could have been doing something worthwhile. Quote Link to comment Share on other sites More sharing options...
Penny Farthing Posted June 10, 2019 Report Share Posted June 10, 2019 On 08/06/2019 at 19:47, Major Cunt said: What are you up to this evening Pen? Also could you clarify your gender (old school definition i.e male/female), I can't keep up with the current gender trends. Ta Oh I forgot .. if you get cunts offering to meet you do not respond .. a few weeks ago I travelled to Dublin to meet a certain cunt at Darkie Kellys in Fishcombe Street. When I got there I asked for the cunt by name and the only response I got was from a one-legged dwarf in a wheelchair who said "Yiz come to the wrong place!", and then muttered something about being a personal trainer and needing to go to Phoenix Park to meet a client and then promptly wheeled himself out of the pub .. out of interest the backrest of his wheelchair had a picture of a WW2 German tank on it. 1 Quote Link to comment Share on other sites More sharing options...
Cuntybaws Posted June 10, 2019 Report Share Posted June 10, 2019 3 hours ago, Glowworm said: How many angels can dance on the head of a pin? Assuming that each angel contains at least one bit of information (fallen / not fallen), and that the point of the pin is a sphere of diameter of an Ångström (R=10exp-10 m) and has a total mass of M=9.5*10exp-29 kilograms (equivalent to that of one iron atom), we can use the Bekenstein bound on information to calculate an upper bound on the angel density. In a system of diameter D and mass M, less than kDM distinguishable bits can exist, where k=2.57686*10exp43 bits/meter kg. This gives us a bound of just 2.448*10exp5 angels, far below the Schewe bound. Note that this does not take the mass of angels into account. A finite angel mass-energy would increase the possible information density significantly. If each angel has a mass m, then the Bekenstein bound gives us N<kD(M+Nm). Beyond mcrit>1/kD ¼3.8807*10exp-34 kg this produces an unbounded maximal angel density as each angel contributes enough mass-energy to allow the information of an extra angel to move in, and so on. However, if angels have mass, then the point of the pin will collapse into a black hole if c2R/2G< Nm (here we ignore the mass of the iron atom at the tip).For angels of human weight (80 kg), we get a limit of 4.2089*10exp14 angels. The maximal mass of any angel amenable to dance on the pin is 3.3671*10exp16 kg; at this point there is only room for a single angel. The picture that emerges is that, for low angel masses, the number is bounded by the Bekenstein bound, and increases hyperbolically as mcrit is approached. However, the black hole bound decreases and the two bounds cross at mmax=1/(4GkM/cexp2+kD), very slightly below mcrit. This corresponds to the maximal angel density of Nmax=8.6766*10exp49 angels. Quantum Gravity Treatment of the Angel Density Problem 3 Quote Link to comment Share on other sites More sharing options...
Jiggerycock Posted June 10, 2019 Report Share Posted June 10, 2019 21 minutes ago, Cuntybaws said: Assuming that each angel contains at least one bit of information (fallen / not fallen), and that the point of the pin is a sphere of diameter of an Ångström (R=10exp-10 m) and has a total mass of M=9.5*10exp-29 kilograms (equivalent to that of one iron atom), we can use the Bekenstein bound on information to calculate an upper bound on the angel density. In a system of diameter D and mass M, less than kDM distinguishable bits can exist, where k=2.57686*10exp43 bits/meter kg. This gives us a bound of just 2.448*10exp5 angels, far below the Schewe bound. Note that this does not take the mass of angels into account. A finite angel mass-energy would increase the possible information density significantly. If each angel has a mass m, then the Bekenstein bound gives us N<kD(M+Nm). Beyond mcrit>1/kD ¼3.8807*10exp-34 kg this produces an unbounded maximal angel density as each angel contributes enough mass-energy to allow the information of an extra angel to move in, and so on. However, if angels have mass, then the point of the pin will collapse into a black hole if c2R/2G< Nm (here we ignore the mass of the iron atom at the tip).For angels of human weight (80 kg), we get a limit of 4.2089*10exp14 angels. The maximal mass of any angel amenable to dance on the pin is 3.3671*10exp16 kg; at this point there is only room for a single angel. The picture that emerges is that, for low angel masses, the number is bounded by the Bekenstein bound, and increases hyperbolically as mcrit is approached. However, the black hole bound decreases and the two bounds cross at mmax=1/(4GkM/cexp2+kD), very slightly below mcrit. This corresponds to the maximal angel density of Nmax=8.6766*10exp49 angels. Quantum Gravity Treatment of the Angel Density Problem Oooh you sexy fuckah! You had me at 'Angstrom' - even put an umlaut on it you little tease! Quote Link to comment Share on other sites More sharing options...
Major Cunt Posted June 10, 2019 Report Share Posted June 10, 2019 3 hours ago, Glowworm said: Oh I forgot .. if you get cunts offering to meet you do not respond .. a few weeks ago I travelled to Dublin to meet a certain cunt at Darkie Kellys in Fishcombe Street. When I got there I asked for the cunt by name and the only response I got was from a one-legged dwarf in a wheelchair who said "Yiz come to the wrong place!", and then muttered something about being a personal trainer and needing to go to Phoenix Park to meet a client and then promptly wheeled himself out of the pub .. out of interest the backrest of his wheelchair had a picture of a WW2 German tank on it. I heard Judge had a similar rendezvous at a restaurant, the rest of the details I'm unfamiliar with apart from it all being a giant stitch up, am I misinformed? I'm chuckling over Panz doing the same! Quote Link to comment Share on other sites More sharing options...
camberwell gypsy Posted June 10, 2019 Report Share Posted June 10, 2019 1 hour ago, Cuntybaws said: Assuming that each angel contains at least one bit of information (fallen / not fallen), and that the point of the pin is a sphere of diameter of an Ångström (R=10exp-10 m) and has a total mass of M=9.5*10exp-29 kilograms (equivalent to that of one iron atom), we can use the Bekenstein bound on information to calculate an upper bound on the angel density. In a system of diameter D and mass M, less than kDM distinguishable bits can exist, where k=2.57686*10exp43 bits/meter kg. This gives us a bound of just 2.448*10exp5 angels, far below the Schewe bound. Note that this does not take the mass of angels into account. A finite angel mass-energy would increase the possible information density significantly. If each angel has a mass m, then the Bekenstein bound gives us N<kD(M+Nm). Beyond mcrit>1/kD ¼3.8807*10exp-34 kg this produces an unbounded maximal angel density as each angel contributes enough mass-energy to allow the information of an extra angel to move in, and so on. However, if angels have mass, then the point of the pin will collapse into a black hole if c2R/2G< Nm (here we ignore the mass of the iron atom at the tip).For angels of human weight (80 kg), we get a limit of 4.2089*10exp14 angels. The maximal mass of any angel amenable to dance on the pin is 3.3671*10exp16 kg; at this point there is only room for a single angel. The picture that emerges is that, for low angel masses, the number is bounded by the Bekenstein bound, and increases hyperbolically as mcrit is approached. However, the black hole bound decreases and the two bounds cross at mmax=1/(4GkM/cexp2+kD), very slightly below mcrit. This corresponds to the maximal angel density of Nmax=8.6766*10exp49 angels. Quantum Gravity Treatment of the Angel Density Problem 1 Quote Link to comment Share on other sites More sharing options...
Last Cunt Standing Posted June 10, 2019 Report Share Posted June 10, 2019 1 hour ago, Cuntybaws said: Assuming that each angel contains at least one bit of information (fallen / not fallen), and that the point of the pin is a sphere of diameter of an Ångström (R=10exp-10 m) and has a total mass of M=9.5*10exp-29 kilograms (equivalent to that of one iron atom), we can use the Bekenstein bound on information to calculate an upper bound on the angel density. In a system of diameter D and mass M, less than kDM distinguishable bits can exist, where k=2.57686*10exp43 bits/meter kg. This gives us a bound of just 2.448*10exp5 angels, far below the Schewe bound. Note that this does not take the mass of angels into account. A finite angel mass-energy would increase the possible information density significantly. If each angel has a mass m, then the Bekenstein bound gives us N<kD(M+Nm). Beyond mcrit>1/kD ¼3.8807*10exp-34 kg this produces an unbounded maximal angel density as each angel contributes enough mass-energy to allow the information of an extra angel to move in, and so on. However, if angels have mass, then the point of the pin will collapse into a black hole if c2R/2G< Nm (here we ignore the mass of the iron atom at the tip).For angels of human weight (80 kg), we get a limit of 4.2089*10exp14 angels. The maximal mass of any angel amenable to dance on the pin is 3.3671*10exp16 kg; at this point there is only room for a single angel. The picture that emerges is that, for low angel masses, the number is bounded by the Bekenstein bound, and increases hyperbolically as mcrit is approached. However, the black hole bound decreases and the two bounds cross at mmax=1/(4GkM/cexp2+kD), very slightly below mcrit. This corresponds to the maximal angel density of Nmax=8.6766*10exp49 angels. Quantum Gravity Treatment of the Angel Density Problem Yeah, but as the Thomist assumption of non-overlap is shite, this calculation is not accurate. Quote Link to comment Share on other sites More sharing options...
Guest Betterthanyou Posted June 10, 2019 Report Share Posted June 10, 2019 1 hour ago, Cuntybaws said: Assuming that each angel contains at least one bit of information (fallen / not fallen), and that the point of the pin is a sphere of diameter of an Ångström (R=10exp-10 m) and has a total mass of M=9.5*10exp-29 kilograms (equivalent to that of one iron atom), we can use the Bekenstein bound on information to calculate an upper bound on the angel density. In a system of diameter D and mass M, less than kDM distinguishable bits can exist, where k=2.57686*10exp43 bits/meter kg. This gives us a bound of just 2.448*10exp5 angels, far below the Schewe bound. Note that this does not take the mass of angels into account. A finite angel mass-energy would increase the possible information density significantly. If each angel has a mass m, then the Bekenstein bound gives us N<kD(M+Nm). Beyond mcrit>1/kD ¼3.8807*10exp-34 kg this produces an unbounded maximal angel density as each angel contributes enough mass-energy to allow the information of an extra angel to move in, and so on. However, if angels have mass, then the point of the pin will collapse into a black hole if c2R/2G< Nm (here we ignore the mass of the iron atom at the tip).For angels of human weight (80 kg), we get a limit of 4.2089*10exp14 angels. The maximal mass of any angel amenable to dance on the pin is 3.3671*10exp16 kg; at this point there is only room for a single angel. The picture that emerges is that, for low angel masses, the number is bounded by the Bekenstein bound, and increases hyperbolically as mcrit is approached. However, the black hole bound decreases and the two bounds cross at mmax=1/(4GkM/cexp2+kD), very slightly below mcrit. This corresponds to the maximal angel density of Nmax=8.6766*10exp49 angels. Quantum Gravity Treatment of the Angel Density Problem Occam's razor would of done the trick. Quote Link to comment Share on other sites More sharing options...
Guest Betterthanyou Posted June 10, 2019 Report Share Posted June 10, 2019 5 hours ago, Glowworm said: How many angels can dance on the head of a pin? First look at your watch or whatever you use to tell the time and note down what time it is. Then and check the name of the site, go and look at all the cunts who post on here and look at their profiles, then go and look at some of their posts and check back on their profiles. You should by then realise that most if not all of them are not who or what they say they are. After doing this check the time and you should come to the realisation that you have just wasted several hours doing something that will not improve your life in any way .. time that you could have been doing something worthwhile. Holy fuck the glowworn has turned (no pun intended), about time, good for you. Quote Link to comment Share on other sites More sharing options...
Penny Farthing Posted June 10, 2019 Report Share Posted June 10, 2019 3 hours ago, Cuntybaws said: Assuming that each angel contains at least one bit of information (fallen / not fallen), and that the point of the pin is a sphere of diameter of an Ångström (R=10exp-10 m) and has a total mass of M=9.5*10exp-29 kilograms (equivalent to that of one iron atom), we can use the Bekenstein bound on information to calculate an upper bound on the angel density. In a system of diameter D and mass M, less than kDM distinguishable bits can exist, where k=2.57686*10exp43 bits/meter kg. This gives us a bound of just 2.448*10exp5 angels, far below the Schewe bound. Note that this does not take the mass of angels into account. A finite angel mass-energy would increase the possible information density significantly. If each angel has a mass m, then the Bekenstein bound gives us N<kD(M+Nm). Beyond mcrit>1/kD ¼3.8807*10exp-34 kg this produces an unbounded maximal angel density as each angel contributes enough mass-energy to allow the information of an extra angel to move in, and so on. However, if angels have mass, then the point of the pin will collapse into a black hole if c2R/2G< Nm (here we ignore the mass of the iron atom at the tip).For angels of human weight (80 kg), we get a limit of 4.2089*10exp14 angels. The maximal mass of any angel amenable to dance on the pin is 3.3671*10exp16 kg; at this point there is only room for a single angel. The picture that emerges is that, for low angel masses, the number is bounded by the Bekenstein bound, and increases hyperbolically as mcrit is approached. However, the black hole bound decreases and the two bounds cross at mmax=1/(4GkM/cexp2+kD), very slightly below mcrit. This corresponds to the maximal angel density of Nmax=8.6766*10exp49 angels. Quantum Gravity Treatment of the Angel Density Problem I could not have put it better myself have a like. Quote Link to comment Share on other sites More sharing options...
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