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  • GIN1138GIN1138 Posts: 22,293
    RobD said:

    Pulpstar said:
    Apparently technological solutions work just fine between the UK and NI, but are utterly hopeless between NI and Ireland.
    ;)
  • volcanopetevolcanopete Posts: 2,078
    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.
  • Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    bridge.
    S.

    I'd guesstimate 1kW, 2 square metres, 30%, 150 billion metres times 4 pi, divide by 9 times ten to the 16, but I think that would give the wrong answer.

    Let's see - 1kW for 2 square metres, implying 500W/square metre power output, implying 1.5kW/sq metre at Earth's distance from Sun, which would be 6 x pi x 10^11 sq metres = about 1.9 x 10^12 square metres gives 2.8 x 10^15 W .
    Divide by 3^8, twice, gives about 0.03kg/s

    Which I know for a fact is wrong. Where did I go wrong?
    My effort is an order of magnitude out. Not bad for a PPEist.

    Say the umbrella is 1m across which gives it a surface area of (U) = 1/4 PI m
    The surface area of a sphere 1.5 * 10^10 km in radius is (S) 4 * 2.25 * 10^20 PI m

    Total umbrellas required to capture all the emitted energy at that distance = (S)/(U) = 3.6 * 10^21

    Each umbrella delivers 1kW to the machine, let's say the panel has 10% efficiency, so each is capturing 10^4 W

    Total energy output (e) from Sun is therefore 3.6 * 10^25

    e = mc^2 so divding through by 9 * 10^16 gives m = 4 * 10^8 kg/s
    I’m wincing slightly at the units, but I’m unusually sensitive to such things. You have the intensity of sunlight a power of ten too big (but estimated perfectly reasonably) so your final answer should be ten times too much, not ten times too small. Something went wrong with the area.
  • Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    So - estimate the amount of energy needed to run a washing machine, then estimate the area covered by an umbrella and the efficiency of the solar cells on it, then the surface area of a sphere at the Earth's distance from the Sun, then divide by c^2? (omitting atmosphere loss and angular tilt of umbrella at whatever latitude you're at; effectively assuming midday at the equator)

    Impressive for the estimate accuracy at each stage to be good enough for a factor of two.

    I'd guesstimate 1kW, 2 square metres, 30%, 150 billion metres times 4 pi, divide by 9 times ten to the 16, but I think that would give the wrong answer.

    Let's see - 1kW for 2 square metres, implying 500W/square metre power output, implying 1.5kW/sq metre at Earth's distance from Sun, which would be 6 x pi x 10^11 sq metres = about 1.9 x 10^12 square metres gives 2.8 x 10^15 W .
    Divide by 3^8, twice, gives about 0.03kg/s

    Which I know for a fact is wrong. Where did I go wrong?
    Pretty good up to the point where you forgot to square the radius before multiplying by four pi.
    Ah, bugger.
    So I multiply by 1.5 x 10^11 to get about 4.5 x 10^9 kg or about 4.5 x 10^6 tonnes or 4.5 million tonnes per second... that's not far off!
    Have a place at Cambridge!
  • Tissue_PriceTissue_Price Posts: 9,039

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    bridge.
    S.

    I'd guesstimate 1kW, 2 square metres, 30%, 150 billion metres times 4 pi, divide by 9 times ten to the 16, but I think that would give the wrong answer.

    Let's see - 1kW for 2 square metres, implying 500W/square metre power output, implying 1.5kW/sq metre at Earth's distance from Sun, which would be 6 x pi x 10^11 sq metres = about 1.9 x 10^12 square metres gives 2.8 x 10^15 W .
    Divide by 3^8, twice, gives about 0.03kg/s

    Which I know for a fact is wrong. Where did I go wrong?
    My effort is an order of magnitude out. Not bad for a PPEist.

    Say the umbrella is 1m across which gives it a surface area of (U) = 1/4 PI m
    The surface area of a sphere 1.5 * 10^10 km in radius is (S) 4 * 2.25 * 10^20 PI m

    Total umbrellas required to capture all the emitted energy at that distance = (S)/(U) = 3.6 * 10^21

    Each umbrella delivers 1kW to the machine, let's say the panel has 10% efficiency, so each is capturing 10^4 W

    Total energy output (e) from Sun is therefore 3.6 * 10^25

    e = mc^2 so divding through by 9 * 10^16 gives m = 4 * 10^8 kg/s
    I’m wincing slightly at the units, but I’m unusually sensitive to such things. You have the intensity of sunlight a power of ten too big (but estimated perfectly reasonably) so your final answer should be ten times too much, not ten times too small. Something went wrong with the area.
    Sorry, I did edit them to m^2! Andy Cooke has pointed out my error re the AU.
  • Tissue_PriceTissue_Price Posts: 9,039

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    So - estimate the amount of energy needed to run a washing machine, then estimate the area covered by an umbrella and the efficiency of the solar cells on it, then the surface area of a sphere at the Earth's distance from the Sun, then divide by c^2? (omitting atmosphere loss and angular tilt of umbrella at whatever latitude you're at; effectively assuming midday at the equator)

    Impressive for the estimate accuracy at each stage to be good enough for a factor of two.

    I'd guesstimate 1kW, 2 square metres, 30%, 150 billion metres times 4 pi, divide by 9 times ten to the 16, but I think that would give the wrong answer.

    Let's see - 1kW for 2 square metres, implying 500W/square metre power output, implying 1.5kW/sq metre at Earth's distance from Sun, which would be 6 x pi x 10^11 sq metres = about 1.9 x 10^12 square metres gives 2.8 x 10^15 W .
    Divide by 3^8, twice, gives about 0.03kg/s

    Which I know for a fact is wrong. Where did I go wrong?
    Pretty good up to the point where you forgot to square the radius before multiplying by four pi.
    Ah, bugger.
    So I multiply by 1.5 x 10^11 to get about 4.5 x 10^9 kg or about 4.5 x 10^6 tonnes or 4.5 million tonnes per second... that's not far off!
    Have a place at Cambridge!
    Commiserations.
  • rottenboroughrottenborough Posts: 62,778

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    So - estimate the amount of energy needed to run a washing machine, then estimate the area covered by an umbrella and the efficiency of the solar cells on it, then the surface area of a sphere at the Earth's distance from the Sun, then divide by c^2? (omitting atmosphere loss and angular tilt of umbrella at whatever latitude you're at; effectively assuming midday at the equator)

    Impressive for the estimate accuracy at each stage to be good enough for a factor of two.

    I'd guesstimate 1kW, 2 square metres, 30%, 150 billion metres times 4 pi, divide by 9 times ten to the 16, but I think that would give the wrong answer.

    Let's see - 1kW for 2 square metres, implying 500W/square metre power output, implying 1.5kW/sq metre at Earth's distance from Sun, which would be 6 x pi x 10^11 sq metres = about 1.9 x 10^12 square metres gives 2.8 x 10^15 W .
    Divide by 3^8, twice, gives about 0.03kg/s

    Which I know for a fact is wrong. Where did I go wrong?
    Pretty good up to the point where you forgot to square the radius before multiplying by four pi.
    Ah, bugger.
    So I multiply by 1.5 x 10^11 to get about 4.5 x 10^9 kg or about 4.5 x 10^6 tonnes or 4.5 million tonnes per second... that's not far off!
    I fear to trend into this, as Nuffield physics left me baffled enough at 18, when I had more brain cells.

    But, isn't approx half the earth in nightime?
  • Fysics_TeacherFysics_Teacher Posts: 6,285
    edited October 2018

    Anorak said:

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    Is that solved by (say) a 1kW machine needing 1000 J/s so the sun loses 1000/c2 per second to power one machine?

    Or did you want the total mass loss, i.e. you estimate the proportion of a sphere around the sun covered by one umbrella?

    Or is this why I did engineering not physics?
    Total mass loss, yes.

    The student concerned read engineering...
    I get about 300,000,000kg?

    (Though I don't think an umbrella-sized panel would be powerful enough to run a washing machine)

    One of those big golfing ones pehaps. I nicked the question from someone else, and it does its job.
  • stodgestodge Posts: 13,910
    Sean_F said:


    As a counterfactual, if Hilary Clinton had won in 2016, the Republicans would be on course to gain about 10 Senate seats.

    I'm less convinced. Clinton's Democrats gained seats in the 1998 midterms as did Bush's Republicans in 2002. You can of course argue the 2002 election was heavily influenced by the events of the preceding year.

    My dim recollection was the GOP was in deep trouble in the summer of 2001 and heading for what I believe is termed in parts of the north as a "shellacking".

    As we don't know for example how Korea would have developed with an HRC Presidency it's impossible to argue the GOP would have bounced back so quickly from a third General Election defeat.

  • rottenboroughrottenborough Posts: 62,778

    Anorak said:

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    Is that solved by (say) a 1kW machine needing 1000 J/s so the sun loses 1000/c2 per second to power one machine?

    Or did you want the total mass loss, i.e. you estimate the proportion of a sphere around the sun covered by one umbrella?

    Or is this why I did engineering not physics?
    Total mass loss, yes.

    The student concerned read engineering...
    I get about 300,000,000kg?

    (Though I don't think an umbrella-sized panel would be powerful enough to run a washing machine)
    When you have all solved this, can you have a look at the NI border question and see if there's a fix? :lol:
  • RobDRobD Posts: 59,936



    Ah, bugger.
    So I multiply by 1.5 x 10^11 to get about 4.5 x 10^9 kg or about 4.5 x 10^6 tonnes or 4.5 million tonnes per second... that's not far off!

    I fear to trend into this, as Nuffield physics left me baffled enough at 18, when I had more brain cells.

    But, isn't approx half the earth in nightime?
    You are wondering about the surface area bit? If you have a measurement of the Sun's radiation at the location of the Earth, the total radiation emitted by the Sun is just that multiplied by ratio of the surface area of the Sun and the surface area of a sphere of radius equal to the Earth-Sun distance.
  • stodgestodge Posts: 13,910

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

  • NemtynakhtNemtynakht Posts: 2,329
    theProle said:

    Anazina said:

    tpfkar said:

    Pulpstar said:

    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    Would they have been going at normal walking speed though?
    No. It was much slower with frequent stops as the march funnelled into narrower sections of the route such as Pall Mall.
    Yes I think the walking speed would be closer to 2 mph, a walk of 4 mph is going somewhere very quickly and in a hurry. I've never seen any march get remotely close to 4 mph.
    I agree, but the slower the walking speed, the lower the attendance in those calculations, and it already seems Alistair's total is well under what others are estimating - so I wonder if that really is a robust way of estimating it?
    The formula is correct: the issue with any estimate is what values you assume for each term. It also assumes all the marchers went past this point.
    God this is a dismal science. You’d have thought by now that someone would have devised a piece of software that could count the size of a crowd within a 100 or so MOE from a helicopter picture. We can put a man on the moon, but we still can’t count a crowd.
    This blogpost is very interesting on the topic of counting crowds.

    https://ahdinnaeken.wordpress.com/2018/05/06/delusion-of-the-35000-indy-marchers/

    I forget where I read it now, but someone has done a sqm estimate of the area occupied by the marchers, and reckoned that to get the 700k claimed you'd be up to 5-6 people sqm (total non starter), i.e. the actual size was probably more like 100k.
    Anyone got any decent helicopter photos of the march, and we can conclusively prove the point.
    I think Alastair first calculation is good apart from the speed, and I agree with the adjusted speed at about 2mph. The only thing I would add is that there may have been a lot of supporters in and around who did not necessarily march the whole time, so I would put the number between 100 and 125k
  • SandyRentoolSandyRentool Posts: 22,044

    theProle said:

    Anyone got any decent helicopter photos of the march, and we can conclusively prove the point.

    I don't think there's one that shows all the people in one shot. The densest section before the march started went up to Marble Arch which isn't shown on any of the aerial shots and others bypassed the start or went straight to Parliament Square.
    "The densest section..." Is that where all the Tim Nice-But-Dim types set off from?
  • sarissasarissa Posts: 1,993
    Nigelb said:

    notme said:

    In the event of No Deal I expect a general strike and marches of at least 20 million.

    I’d expect Drumhead trials for those Leavers who said it would be easy/WTO was Project Fear.


    I think worst case scenario is massive disruption for a period of months, and then some lost growth, possible recession. Capitalism has an amazing way of finding away....
    Freudian slip ?

    Capitalism does indeed usually find a way profit, but for some (see the interviews with scientists of the Crick Institute this morning), that might well be outside of the UK.
    Fixed that for you
  • Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    de you're at; effectively assuming midday at the equator)

    Impressive for the estimate accuracy at each stage to be good enough for a factor of two.

    I'd guesstimate 1kW, 2 square metres, 30%, 150 billion metres times 4 pi, divide by 9 times ten to the 16, but I think that would give the wrong answer.

    Let's see - 1kW for 2 square metres, implying 500W/square metre power output, implying 1.5kW/sq metre at Earth's distance from Sun, which would be 6 x pi x 10^11 sq metres = about 1.9 x 10^12 square metres gives 2.8 x 10^15 W .
    Divide by 3^8, twice, gives about 0.03kg/s

    Which I know for a fact is wrong. Where did I go wrong?
    Pretty good up to the point where you forgot to square the radius before multiplying by four pi.
    Ah, bugger.
    So I multiply by 1.5 x 10^11 to get about 4.5 x 10^9 kg or about 4.5 x 10^6 tonnes or 4.5 million tonnes per second... that's not far off!
    I fear to trend into this, as Nuffield physics left me baffled enough at 18, when I had more brain cells.

    But, isn't approx half the earth in nightime?
    Nuffield Physics included estimation as a question type.

    Mind you, I did Nuffield Chemistry A-level as well and couldn’t tell you a thing about it now.

    Approx half the Earth is in night time yes, but it doesn’t matter in this case. Can you see why?

    Sorry: I was going into teacher mode there...
  • anothernickanothernick Posts: 3,591
    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
  • stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    Two stories: for supporters the idea that their opponents are trying to minimise what happed an so must be worried about it; for opponents the idea that the marchers have been caught out exaggerating their level of support to make it seem more significant.

    Both sides will be satisfied.
  • Tissue_PriceTissue_Price Posts: 9,039
    These resignations could be tricky for Bercow.
    https://twitter.com/BBCVickiYoung/status/1054718690804015105
  • Richard_NabaviRichard_Nabavi Posts: 30,821
    edited October 2018

    Anorak said:

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    Is that solved by (say) a 1kW machine needing 1000 J/s so the sun loses 1000/c2 per second to power one machine?

    Or did you want the total mass loss, i.e. you estimate the proportion of a sphere around the sun covered by one umbrella?

    Or is this why I did engineering not physics?
    Total mass loss, yes.

    The student concerned read engineering...
    I get about 300,000,000kg?

    (Though I don't think an umbrella-sized panel would be powerful enough to run a washing machine)
    When you have all solved this, can you have a look at the NI border question and see if there's a fix? :lol:
    I think you need to start by estimating the ratio of the surface area of a bowler hat to the annual volume of whiskey smuggled through the border.
  • david_herdsondavid_herdson Posts: 17,751

    Anorak said:

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    Is that solved by (say) a 1kW machine needing 1000 J/s so the sun loses 1000/c2 per second to power one machine?

    Or did you want the total mass loss, i.e. you estimate the proportion of a sphere around the sun covered by one umbrella?

    Or is this why I did engineering not physics?
    Total mass loss, yes.

    The student concerned read engineering...
    I get about 300,000,000kg?

    (Though I don't think an umbrella-sized panel would be powerful enough to run a washing machine)

    One of those big golfing ones pehaps. I nicked the question from someone else, and it does its job.
    I conveniently estimated the area of the umbrella to be exactly one square metre - perhaps the reason why my answer was a little less than Andy Cooke's (though it was posted earlier!). I'm an order of magnitude out somewhere, though. My fault for not writing down my working.
  • williamglennwilliamglenn Posts: 51,752

    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    Two stories: for supporters the idea that their opponents are trying to minimise what happed an so must be worried about it; for opponents the idea that the marchers have been caught out exaggerating their level of support to make it seem more significant.

    Both sides will be satisfied.
    We can solve it with a referendum: Were you on the People's Vote march? Yes/No.
  • Andy_CookeAndy_Cooke Posts: 5,005

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    So - estimate the amount of energy needed to run a washing machine, then estimate the area covered by an umbrella and the efficiency of the solar cells on it, then the surface area of a sphere at the Earth's distance from the Sun, then divide by c^2? (omitting atmosphere loss and angular tilt of umbrella at whatever latitude you're at; effectively assuming midday at the equator)

    Impressive for the estimate accuracy at each stage to be good enough for a factor of two.

    I'd guesstimate 1kW, 2 square metres, 30%, 150 billion metres times 4 pi, divide by 9 times ten to the 16, but I think that would give the wrong answer.

    Let's see - 1kW for 2 square metres, implying 500W/square metre power output, implying 1.5kW/sq metre at Earth's distance from Sun, which would be 6 x pi x 10^11 sq metres = about 1.9 x 10^12 square metres gives 2.8 x 10^15 W .
    Divide by 3^8, twice, gives about 0.03kg/s

    Which I know for a fact is wrong. Where did I go wrong?
    Pretty good up to the point where you forgot to square the radius before multiplying by four pi.
    Ah, bugger.
    So I multiply by 1.5 x 10^11 to get about 4.5 x 10^9 kg or about 4.5 x 10^6 tonnes or 4.5 million tonnes per second... that's not far off!
    Have a place at Cambridge!
    I went to Imperial, as it happens :)
  • Tissue_PriceTissue_Price Posts: 9,039
    If the letters go in against May, I note @david_herdson has been suggesting she might well lose, on the basis that enough MPs would take the chance of a resolution, even though they hadn't wanted to bring things to a head.

    Q: Could she head that off by promising to quit in April?
  • david_herdsondavid_herdson Posts: 17,751

    Anorak said:

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    Is that solved by (say) a 1kW machine needing 1000 J/s so the sun loses 1000/c2 per second to power one machine?

    Or did you want the total mass loss, i.e. you estimate the proportion of a sphere around the sun covered by one umbrella?

    Or is this why I did engineering not physics?
    Total mass loss, yes.

    The student concerned read engineering...
    I get about 300,000,000kg?

    (Though I don't think an umbrella-sized panel would be powerful enough to run a washing machine)
    When you have all solved this, can you have a look at the NI border question and see if there's a fix? :lol:
    Exactness is not a quality best brought to Irish politics. It works best with constructive ambiguity.
  • If the letters go in against May, I note @david_herdson has been suggesting she might well lose, on the basis that enough MPs would take the chance of a resolution, even though they hadn't wanted to bring things to a head.

    Q: Could she head that off by promising to quit in April?

    Quite possibly, although probably a bit vaguer on timing ('summer 2019').
  • PulpstarPulpstar Posts: 78,220
    RobD said:



    Ah, bugger.
    So I multiply by 1.5 x 10^11 to get about 4.5 x 10^9 kg or about 4.5 x 10^6 tonnes or 4.5 million tonnes per second... that's not far off!

    I fear to trend into this, as Nuffield physics left me baffled enough at 18, when I had more brain cells.

    But, isn't approx half the earth in nightime?
    You are wondering about the surface area bit? If you have a measurement of the Sun's radiation at the location of the Earth, the total radiation emitted by the Sun is just that multiplied by ratio of the surface area of the Sun and the surface area of a sphere of radius equal to the Earth-Sun distance.
    Radius = 150,000,000 km, so the surface area is 4 * pi * (150 mill x 1000)^2

    Assume the mean photon wavelength is 562 nm and this https://socratic.org/questions/calculate-the-energy-in-joules-of-a-photon-of-green-light-having-a-wavelength-of bit of cribbing gives us 3.5 x 10^-19 J.

    Which means the sun emits 8 x 10^44 visible light photons/second, I err think.
  • RobDRobD Posts: 59,936
    edited October 2018

    Anorak said:

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    Is that solved by (say) a 1kW machine needing 1000 J/s so the sun loses 1000/c2 per second to power one machine?

    Or did you want the total mass loss, i.e. you estimate the proportion of a sphere around the sun covered by one umbrella?

    Or is this why I did engineering not physics?
    Total mass loss, yes.

    The student concerned read engineering...
    I get about 300,000,000kg?

    (Though I don't think an umbrella-sized panel would be powerful enough to run a washing machine)
    When you have all solved this, can you have a look at the NI border question and see if there's a fix? :lol:
    Exactness is not a quality best brought to Irish politics. It works best with constructive ambiguity.
    Consider a spherical horse border..... :D
  • williamglennwilliamglenn Posts: 51,752

    If the letters go in against May, I note @david_herdson has been suggesting she might well lose, on the basis that enough MPs would take the chance of a resolution, even though they hadn't wanted to bring things to a head.

    Q: Could she head that off by promising to quit in April?

    She can promise to quit after Brexit, then win the vote before calling a second referendum, thus voiding her promise.
  • Anorak said:

    Doing some rough and ready calculations from that sped-up youtube video of the march, I come up with a figure of about 300,000 attendees.

    Yet @Recidivist made it 670,000. How do we bet?
    I'll give you my workings.

    Each line had something like 20 people in it, so far as I can tell. Each line is spaced about a yard apart. That means there's 35,000 or so marchers a mile.

    A normal walking speed is 3-4 miles an hour. The video states that the march took 2 1/2 hours to pass by. Getting rid of all false precision, I get to 300,000.

    I'm very open to other calculations.
    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.
    Is that solved by (say) a 1kW machine needing 1000 J/s so the sun loses 1000/c2 per second to power one machine?

    Or did you want the total mass loss, i.e. you estimate the proportion of a sphere around the sun covered by one umbrella?

    Or is this why I did engineering not physics?
    Total mass loss, yes.

    The student concerned read engineering...
    I get about 300,000,000kg?

    (Though I don't think an umbrella-sized panel would be powerful enough to run a washing machine)

    One of those big golfing ones pehaps. I nicked the question from someone else, and it does its job.
    I conveniently estimated the area of the umbrella to be exactly one square metre - perhaps the reason why my answer was a little less than Andy Cooke's (though it was posted earlier!). I'm an order of magnitude out somewhere, though. My fault for not writing down my working.
    The original student made two mistakes which cancelled out.

    The important point is about thinking very carefully about where calculated data comes from and hoe well we know each of the inputs. It’s also about doing rough and ready order of magnitude calculations so that when you put numbers into a calculator you don’t go badly wrong.

    Have a good afternoon everyone. I’m going out for a walk in the sun.
  • CyclefreeCyclefree Posts: 25,318
    edited October 2018
    Well, why you have all been arguing about how many people went on a march and how many people may or may not have sent in letters to Mr Brady, I decided to ignore my feeble biceps and the tenosynovitis it is suffering from, dosed myself up with painkillers to “away with the fairies “level and, with the help of Eldest Son, who helpfully lugged 850 litres of compost and bark chippings from the car to the garden, spent all weekend planting hundreds of bulbs.

    Simply glorious: the autumn light and sunshine made it memorable. And the pain has not noticeably worsened.

    Only a few more bulbs to go.....

    I am now on squirrel watch.

    But at least I will have something beautiful to look forward to next spring. :)
  • Andy_CookeAndy_Cooke Posts: 5,005



    A good estimation question is one of my staples when conducting mock oxbridge interviews. I once asked “if an umbrella covered in solar panels can run a washing machine, what is the rate of mass loss from the sun?” And got an answer which was within a factor of two. Wasn’t surprised when he got in to Cambridge.

    So - estimate the amount of energy needed to run a washing machine, then estimate the area covered by an umbrella and the efficiency of the solar cells on it, then the surface area of a sphere at the Earth's distance from the Sun, then divide by c^2? (omitting atmosphere loss and angular tilt of umbrella at whatever latitude you're at; effectively assuming midday at the equator)

    Impressive for the estimate accuracy at each stage to be good enough for a factor of two.

    I'd guesstimate 1kW, 2 square metres, 30%, 150 billion metres times 4 pi, divide by 9 times ten to the 16, but I think that would give the wrong answer.

    Let's see - 1kW for 2 square metres, implying 500W/square metre power output, implying 1.5kW/sq metre at Earth's distance from Sun, which would be 6 x pi x 10^11 sq metres = about 1.9 x 10^12 square metres gives 2.8 x 10^15 W .
    Divide by 3^8, twice, gives about 0.03kg/s

    Which I know for a fact is wrong. Where did I go wrong?
    Pretty good up to the point where you forgot to square the radius before multiplying by four pi.
    Ah, bugger.
    So I multiply by 1.5 x 10^11 to get about 4.5 x 10^9 kg or about 4.5 x 10^6 tonnes or 4.5 million tonnes per second... that's not far off!
    I fear to trend into this, as Nuffield physics left me baffled enough at 18, when I had more brain cells.

    But, isn't approx half the earth in nightime?
    Visualise the Sun giving off energy as if it's pushing out a shell of light, like a bubble growing and becoming more and more diffuse.
    All the energy is concentrated in a bubble with surface area equal to that of the Sun as it is first emitted, spreading out over a greater and greater surface area as it expands outwards into the cosmos.

    By the time the light reaches Earth, that bubble's surface area is spread out over a sphere the size of Earth's orbit around the Sun - a bubble 150 million kilometres in radius. But the same total amount of energy per second; just spread out over that far larger surface area.

    So we calculate the number of square metres of that huge bubble of light and work out what fraction of that bubble's surface area is covered by one umbrella at the same distance.

    Does that help?
  • Cyclefree said:

    Well, why you have all been arguing about how many people went on a march and how many people may or may not have sent in letters to Mr Brady, I decided to ignore my feeble biceps and the tenosynovitis it is suffering from, dosed myself up with painkillers to “away with the fairies “level and, with the help of Eldest Son, who helpfully lugged 850 litres of compost and bark chippings from the car to the garden, spent all weekend planting hundreds of bulbs.

    Simply glorious: the autumn light and sunshine made it memorable. And the pain has not noticeably worsened.

    Only a few more bulbs to go.....

    I am now on squirrel watch.

    Bur at least I will have something beautiful to look forward to next spring. :)

    A row of dead squirrels?
  • david_herdsondavid_herdson Posts: 17,751
    stodge said:

    Sean_F said:


    As a counterfactual, if Hilary Clinton had won in 2016, the Republicans would be on course to gain about 10 Senate seats.

    I'm less convinced. Clinton's Democrats gained seats in the 1998 midterms as did Bush's Republicans in 2002. You can of course argue the 2002 election was heavily influenced by the events of the preceding year.

    My dim recollection was the GOP was in deep trouble in the summer of 2001 and heading for what I believe is termed in parts of the north as a "shellacking".

    As we don't know for example how Korea would have developed with an HRC Presidency it's impossible to argue the GOP would have bounced back so quickly from a third General Election defeat.

    It would have been a third *presidential* defeat but the GOP's record in congressional elections is a lot more impressive. In as far as HRC would have had an effect, I imagine it would depend if hearings into e-mails had got anywhere and perhaps how hearings into *her* SCOTUS nominees had gone. The economy would probably be running nicely, though not as aggressively as under Trump - no big tax cuts - so I think 10 GOP gains is on the high side but certainly the Republicans would have been likely to increase their majority to a comfortable level.
  • PulpstarPulpstar Posts: 78,220
    edited October 2018



    Visualise the Sun giving off energy as if it's pushing out a shell of light, like a bubble growing and becoming more and more diffuse.
    All the energy is concentrated in a bubble with surface area equal to that of the Sun as it is first emitted, spreading out over a greater and greater surface area as it expands outwards into the cosmos.

    By the time the light reaches Earth, that bubble's surface area is spread out over a sphere the size of Earth's orbit around the Sun - a bubble 150 million kilometres in radius. But the same total amount of energy per second; just spread out over that far larger surface area.

    So we calculate the number of square metres of that huge bubble of light and work out what fraction of that bubble's surface area is covered by one umbrella at the same distance.

    Does that help?

    Yes, that's right -

    The difficult/unknown parts of this equation are the panel efficiency, atmospheric loss and photons emitted outside the visible wavelength. Any calculated answer of loss from the given information will be a big underestimate I think.
  • kjhkjh Posts: 11,819

    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    +1
  • Tissue_PriceTissue_Price Posts: 9,039
    edited October 2018
    kjh said:

    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    +1
    The fact that the initial number was clearly a deliberate overestimation (aka a lie, in less Parliamentary terms) also tells its own story.

    But 'twas ever thus with marches, regardless of their political persuasion. Maybe we need a MarchCheck website which provides independent estimates for all of them...
  • RobDRobD Posts: 59,936
    Pulpstar said:



    Visualise the Sun giving off energy as if it's pushing out a shell of light, like a bubble growing and becoming more and more diffuse.
    All the energy is concentrated in a bubble with surface area equal to that of the Sun as it is first emitted, spreading out over a greater and greater surface area as it expands outwards into the cosmos.

    By the time the light reaches Earth, that bubble's surface area is spread out over a sphere the size of Earth's orbit around the Sun - a bubble 150 million kilometres in radius. But the same total amount of energy per second; just spread out over that far larger surface area.

    So we calculate the number of square metres of that huge bubble of light and work out what fraction of that bubble's surface area is covered by one umbrella at the same distance.

    Does that help?

    Yes, that's right -

    The difficult/unknown parts of this equation are the panel efficiency, atmospheric loss and photons emitted outside the visible wavelength. Any calculated answer of loss from the given information will be a big underestimate I think.
    Atmosphere is pretty much transparent at optical wavelengths, although I’m not far how blue/red solar panels go. Efficiency will be the biggest loss, I think.
  • anothernickanothernick Posts: 3,591

    Cyclefree said:

    Well, why you have all been arguing about how many people went on a march and how many people may or may not have sent in letters to Mr Brady, I decided to ignore my feeble biceps and the tenosynovitis it is suffering from, dosed myself up with painkillers to “away with the fairies “level and, with the help of Eldest Son, who helpfully lugged 850 litres of compost and bark chippings from the car to the garden, spent all weekend planting hundreds of bulbs.

    Simply glorious: the autumn light and sunshine made it memorable. And the pain has not noticeably worsened.

    Only a few more bulbs to go.....

    I am now on squirrel watch.

    Bur at least I will have something beautiful to look forward to next spring. :)

    A row of dead squirrels?
    Or maybe a dead Brexit?
  • PulpstarPulpstar Posts: 78,220
    RobD said:


    Atmosphere is pretty much transparent at optical wavelengths

    Lol, is it always clear in California :p ?
  • RobDRobD Posts: 59,936
    Pulpstar said:

    RobD said:


    Atmosphere is pretty much transparent at optical wavelengths

    Lol, is it always clear in California :p ?
    Had my astronomer hat on. Good point about clouds! :p
  • kjhkjh Posts: 11,819

    Cyclefree said:

    Well, why you have all been arguing about how many people went on a march and how many people may or may not have sent in letters to Mr Brady, I decided to ignore my feeble biceps and the tenosynovitis it is suffering from, dosed myself up with painkillers to “away with the fairies “level and, with the help of Eldest Son, who helpfully lugged 850 litres of compost and bark chippings from the car to the garden, spent all weekend planting hundreds of bulbs.

    Simply glorious: the autumn light and sunshine made it memorable. And the pain has not noticeably worsened.

    Only a few more bulbs to go.....

    I am now on squirrel watch.

    Bur at least I will have something beautiful to look forward to next spring. :)

    A row of dead squirrels?
    :)
  • david_herdsondavid_herdson Posts: 17,751
    RobD said:

    Pulpstar said:



    Visualise the Sun giving off energy as if it's pushing out a shell of light, like a bubble growing and becoming more and more diffuse.
    All the energy is concentrated in a bubble with surface area equal to that of the Sun as it is first emitted, spreading out over a greater and greater surface area as it expands outwards into the cosmos.

    By the time the light reaches Earth, that bubble's surface area is spread out over a sphere the size of Earth's orbit around the Sun - a bubble 150 million kilometres in radius. But the same total amount of energy per second; just spread out over that far larger surface area.

    So we calculate the number of square metres of that huge bubble of light and work out what fraction of that bubble's surface area is covered by one umbrella at the same distance.

    Does that help?

    Yes, that's right -

    The difficult/unknown parts of this equation are the panel efficiency, atmospheric loss and photons emitted outside the visible wavelength. Any calculated answer of loss from the given information will be a big underestimate I think.
    Atmosphere is pretty much transparent at optical wavelengths, although I’m not far how blue/red solar panels go. Efficiency will be the biggest loss, I think.
    Solar panels are not eyes, so aren't 'panel efficiency' and 'photons emitted outside the usable wavelength' elements of the same thing? it doesn't matter whether we can see UV or not; what matters is how well the panel can process it.
  • kjhkjh Posts: 11,819

    kjh said:

    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    +1
    The fact that the initial number was clearly a deliberate overestimation (aka a lie, in less Parliamentary terms) also tells its own story.

    But 'twas ever thus with marches, regardless of their political persuasion. Maybe we need a MarchCheck website which provides independent estimates for all of them...
    Was it? I would expect the organisers to spin the numbers, but there hasn't really been any unbiased contradiction in the media. Obviously CR thinks there were only 5 people on the march. If there was a significant exaggeration they have done a good job in getting away with it.
  • RobDRobD Posts: 59,936

    RobD said:

    Pulpstar said:



    Visualise the Sun giving off energy as if it's pushing out a shell of light, like a bubble growing and becoming more and more diffuse.
    All the energy is concentrated in a bubble with surface area equal to that of the Sun as it is first emitted, spreading out over a greater and greater surface area as it expands outwards into the cosmos.

    By the time the light reaches Earth, that bubble's surface area is spread out over a sphere the size of Earth's orbit around the Sun - a bubble 150 million kilometres in radius. But the same total amount of energy per second; just spread out over that far larger surface area.

    So we calculate the number of square metres of that huge bubble of light and work out what fraction of that bubble's surface area is covered by one umbrella at the same distance.

    Does that help?

    Yes, that's right -

    The difficult/unknown parts of this equation are the panel efficiency, atmospheric loss and photons emitted outside the visible wavelength. Any calculated answer of loss from the given information will be a big underestimate I think.
    Atmosphere is pretty much transparent at optical wavelengths, although I’m not far how blue/red solar panels go. Efficiency will be the biggest loss, I think.
    Solar panels are not eyes, so aren't 'panel efficiency' and 'photons emitted outside the usable wavelength' elements of the same thing? it doesn't matter whether we can see UV or not; what matters is how well the panel can process it.
    Yeah, I’m not sure how blue/red in wavelength terms PV cells go. I had thought it was cantered on the visible wavelengths, at the peak of the Sun’s emission, and where most of the photons get through.
  • Tissue_PriceTissue_Price Posts: 9,039
    4-3 isn't the best start chasing 367, is it?
  • RobDRobD Posts: 59,936
    This thread is officially old news.
  • OllyTOllyT Posts: 5,006
    I
    Sean_F said:

    HYUFD said:

    notme said:

    In the event of No Deal I expect a general strike and marches of at least 20 million.

    I’d expect Drumhead trials for those Leavers who said it would be easy/WTO was Project Fear.


    I think worst case scenario is massive disruption for a period of months, and then some lost growth, possible recession. Capitalism has an amazing way of finding away. I don’t want to leave the single market, it will leave us all a bit porter. But only a bit.
    No, worst case scenario is that AND Scotland voting for independence and Northern Ireland for a United Ireland.

    So the economy could collapse followed by the country breaking up
    Not to mention starving to death and bubonic plague.
    I doubt snide comments are going to reassure many people that No Deal will be fine.
  • OllyTOllyT Posts: 5,006

    Having seen footage of the march, here is my estimate:

    Number of working class voters from Stoke or Sunderland on the march: Zero.


    It would be a pretty safe assumption that the number of working class voters from Stoke or Sunderland would be zero on any sort of march you care to imagine.
  • FoxyFoxy Posts: 48,749
    kjh said:

    kjh said:

    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    +1
    The fact that the initial number was clearly a deliberate overestimation (aka a lie, in less Parliamentary terms) also tells its own story.

    But 'twas ever thus with marches, regardless of their political persuasion. Maybe we need a MarchCheck website which provides independent estimates for all of them...
    Was it? I would expect the organisers to spin the numbers, but there hasn't really been any unbiased contradiction in the media. Obviously CR thinks there were only 5 people on the march. If there was a significant exaggeration they have done a good job in getting away with it.
    The estimates that I have seen came from police and broadcasters, not the organisers. Certainly a very large march/protest and the second biggest in British history.

    It seems that @Casino_Royale and some other Leavers are sufferring cognitive dissonance. They cannot understand why so many ordinary British folk object to the fantasy Brexit that they peddle.
  • RobDRobD Posts: 59,936
    Foxy said:

    kjh said:

    kjh said:

    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    +1
    The fact that the initial number was clearly a deliberate overestimation (aka a lie, in less Parliamentary terms) also tells its own story.

    But 'twas ever thus with marches, regardless of their political persuasion. Maybe we need a MarchCheck website which provides independent estimates for all of them...
    Was it? I would expect the organisers to spin the numbers, but there hasn't really been any unbiased contradiction in the media. Obviously CR thinks there were only 5 people on the march. If there was a significant exaggeration they have done a good job in getting away with it.
    The estimates that I have seen came from police and broadcasters, not the organisers. Certainly a very large march/protest and the second biggest in British history.

    It seems that @Casino_Royale and some other Leavers are sufferring cognitive dissonance. They cannot understand why so many ordinary British folk object to the fantasy Brexit that they peddle.
    Any evidence to suggest those marching were ordinary folk? The ordinary person wasn’t marching at the weekend.
  • rpjsrpjs Posts: 3,787
    RobD said:

    Pulpstar said:
    Apparently technological solutions work just fine between the UK and NI, but are utterly hopeless between NI and Ireland.
    It's a lot easier to police a maritime border with a small number of ports of entry, and radar to detect vessels attempting to avoid them, than a hundreds of miles long land border with literally hundreds of roads crossing it.

    Yes, during the Troubles the UK closed most of the road crossings and used patrols and other technology to deter border crossings away from the designated crossing places but that's exactly the hard border we're trying to avoid and it wasn't very successful anyway.
  • FoxyFoxy Posts: 48,749
    RobD said:

    Foxy said:

    kjh said:

    kjh said:

    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    +1
    The fact that the initial number was clearly a deliberate overestimation (aka a lie, in less Parliamentary terms) also tells its own story.

    But 'twas ever thus with marches, regardless of their political persuasion. Maybe we need a MarchCheck website which provides independent estimates for all of them...
    Was it? I would expect the organisers to spin the numbers, but there hasn't really been any unbiased contradiction in the media. Obviously CR thinks there were only 5 people on the march. If there was a significant exaggeration they have done a good job in getting away with it.
    The estimates that I have seen came from police and broadcasters, not the organisers. Certainly a very large march/protest and the second biggest in British history.

    It seems that @Casino_Royale and some other Leavers are sufferring cognitive dissonance. They cannot understand why so many ordinary British folk object to the fantasy Brexit that they peddle.
    Any evidence to suggest those marching were ordinary folk? The ordinary person wasn’t marching at the weekend.
    The ones around me were pretty ordinary. It may be harder to tell from so far away as the USA.
  • RobDRobD Posts: 59,936
    Foxy said:

    RobD said:

    Foxy said:

    kjh said:

    kjh said:

    stodge said:

    I very much like Peter Dowd's response to the Maybot's claim that the Brexit deal is "95% done" that the Titanic was "95% succesful" too.A little dry humour hits the spot far better than abuse.

    Unfortunately dry humour is no match for the pointless pedantry which pervades Internet discourse these days. Some people are arguing over whether it was 700,000, 100,000 or just 70 people marching last Saturday.

    Who cares ? As those deep meaning philosophers Bewitched once advised "Get A Life".

    The fact that people are still talking about it and trying to minimise the numbers tells its own story.
    +1
    The fact that the initial number was clearly a deliberate overestimation (aka a lie, in less Parliamentary terms) also tells its own story.

    But 'twas ever thus with marches, regardless of their political persuasion. Maybe we need a MarchCheck website which provides independent estimates for all of them...
    Was it? I would expect the organisers to spin the numbers, but there hasn't really been any unbiased contradiction in the media. Obviously CR thinks there were only 5 people on the march. If there was a significant exaggeration they have done a good job in getting away with it.
    The estimates that I have seen came from police and broadcasters, not the organisers. Certainly a very large march/protest and the second biggest in British history.

    It seems that @Casino_Royale and some other Leavers are sufferring cognitive dissonance. They cannot understand why so many ordinary British folk object to the fantasy Brexit that they peddle.
    Any evidence to suggest those marching were ordinary folk? The ordinary person wasn’t marching at the weekend.
    The ones around me were pretty ordinary. It may be harder to tell from so far away as the USA.
    Given the numbers I think by definition they aren’t ordinary.
  • timmo said:

    Being reported that body parts of Kashoggi have now been found..this goes on and on..

    What the call a finger tip search.
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