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I'm here in London, in the middle of the city, and behind me you can see the amazing building at 20 Fenchurch Street.
you can see that phenomenal curved front on it.
and when they were originally putting it together,
during the construction
something mysterious started to happen.
Nearby cars started to melt;
paint was blistering,
a bike seat caught on fire,
someone unexpectedly scorched their lemon.
and that's not even a euphemism.
You're in London,
you're out taking your lemon for a walk.
sunlight is hitting it, but you're not worried.
even though sunlight carries a lot of energy,
not enough is hitting the lemon at once for it to become scorched.
But then
you walk past a reflective building
sunlight is now bouncing off the windows on that building onto your lemon, and it is
still fine. Not enough is hitting it at once.
Only if that building is accidentally curved
in a certain way
will it take sunlight over a large area
and focus in onto this one little lemony point
and then suddenly
there it goes
you've scorched your lemon.
Now I will admit the whole lemon thing is a bit ridiculous
The only citation I could find for it was a Daily Mail article.
But the BBC did confirm that the curved surface on the 20 Fenchurch Street Building
did melt someone's car.
And the Independent Newspaper reported that that fantastic curved surface
focused enough light
to burn this doormat
in a nearby barber shop.
Thankfully, the sun was only in the right position
to hit the curved surface of the buildling
and cause an actual problem
for a couple of weeks of the year
and even then
it was only an issue for a few hours each day
and even then
it was only an issue if it was actually sunny
which, in London,
is not a major concern.
And just to make sure it's completely safe
for that occasional sunny day
the building has since been fitted with a very large sunglass-style monocle, I guess
I mean, it is in London, after all.
Could it have been worse?
Was the 20 Fenchurch Street the worst possible shape to focus sunlight in?
Well, I couldn't find out what shape it actually is,
but I can confirm, having looked at it
for a while
that it's probably not the worst possible shape.
That would be a shape known as the paraboloid.
To make a parabaloid, you start with a parabola
and then you rotate it around the vertical axis.
and there you are,
instant paraboloid.
The great thing about a shape like this
is all the parallel light
that goes straight into it will be focused in
on one focal point.
It's incredibly useful.
If you want to focus radiation,
this is the shape for you.
So, something like a satellite dish
where you want to get lots of incoming radiation
onto one receiver.
Satellite dishes are actually paraboloids
as you can see on my satellite dish here
and if you're concerned that that is a ridiculously big satellite dish
of course, I'm kidding.
That's not my satellite dish.
THAT is my satellite dish.
Hypothetically, the 20 Fenchurch Street could have been worse
but surely, nobody is going to make a massive shiny structure
that is shaped like a paraboloid?
Here in Nottingham, at the Nottingham Playhouse Theatre
There is actually a massive paraboloid sculpture.
Over five metres tall
it's known as the sky mirror
you can see, it's actually not that wildly different to mine
there we go
in fact, on a sunny day
it's even been rumoured
that this thing
will set pigeons on fire.
Yeah, I may have bought some paraboloids
but let me explain.
I'm part of a nerd comedy group called Festival of the Spoken Nerd.
I'm approximately one third
and on our tour with the show "Just for graphs"
I actually have a paraboloid or two in the show itself.
We're actually about to do the final performance on the tour of "Just for Graphs"
We're here in the Queen Mary University of London's People's Palace
which is an amazing theatre
and given it is the last show on the tour
I thought I would pop down to the stage
and show you what we use these paraboloids actually for.
On the right hand side of the stage
I'm putting one paraboloid on a tripod
pointing over towards the left hand side of the stage
where I have a second paraboloid
pointing perfectly back at it.
On the right hand side of the stage
I have an incredibly bright heat lamp
and on the left hand side, I have some flash cotton
which is incredibly flammable
Fine, I'll turn the heat lamp on
oh goodness, right.
In theory, if I get this perfectly in the focal point on this side
it will focus all the light in on the flash cotton
and once it heats up
YES!
Thank you
Thank you for watching my video all about paraboloids
Now as I mentioned, I did do that demonstration
fire in our Spoken Nerd show
"Just for Graphs"
which, by the time you watch this, will be over.
but don't worry, I'm sure we're doing shows somewhere.
in our previous show
"Full Frontal Nerdity".
is available for download
and as a DVD
if you want to see more Spoken Nerd stuff,
just click on, well, actually,
the logo on the screen.
in the background
like there.
Good video this week everybody
we have the building that sets things on fire,
that is a fantastic hook to get people interested
oh, we had the hilarious lemon
the one that magically appears
and disappears
at random
there was satellite dishes
fantastic. Practical application. Good job.
we had the proof for why parabolas all focus
oh wait...
people. We forgot the proof!
Oh, my goodness
there's some very upset viewers, look
ok, look
ok
we will get a proof together
just give me a moment
I have a proof
now, interesting, when most people meet parabolas,
they come across them as quadratics
a plot an an 𝒙² equation
however, parabolas are also intimately linked to the concept of a focal point.
one definition of a parabola
which ignores quadratics completely,
is that it's all the points which are the same distance
from another point
and a line
If you want to, I'll have a link to the example here
so you can have a play with it
and see how that works
but I wasn't happy with that
I wanted to show
starting with an 𝒙² equation
can we prove a parabola
has a focal point?
but for that
it looks like we need
a chalkboard
and some working out
I'm going to start with a standard vertical 𝑦-axis
horizontal 𝒙-axis
and onto this
I'm going to put everyone's standard issue parabola
𝑦=𝒙²
now
we can have a look at what happens
if we have a single beam of light
coming directly down
parallel to the vertical axis
impacts the reflective surface of our parabola
and goes flying off in a new direction
to see what happens
at that point of impact
you're going to have to zoom right in
and I'm talking uncomfortably close
into this point here
well, this really is uncomfortably close
but here we are - the point where it hits our parabola
I'm actually going to take off the arrowhead for now
just to make it a little bit clearer what's going on
and I'm going to draw the line
if it continues to go on straight down and hits the 𝒙-axis
and because we're going to do this in general, we're going to hit the 𝒙-axis at any point
𝒙, which I do apologise, is slightly out of shot
there. We can now approximate the parabola because we don't care about how it curves over there
we only care about the gradient where it impacts
so we can replace that with a tangent
this pink line here
that is the tangent
right at that point where it hits
the line and it is now going to bounce off in an all-new direction.
and so, here comes our light beam, it hits the surface there
and it flies off such that the angle onto, I guess we could say, mirror
it goes on the same angle out, so there it goes off into it's new direction
to calculate the angle it's going on
we need to know initially what angle our mirror is
and we know that for a 𝑦=𝒙² parabola
that the gradient
is the derivative at that point
and the derivative of 𝑦=𝒙²
is 2𝒙
and so the gradient is always twice what the value of 𝒙 happens to be
so, let's label that in vaguely
so for every 1 we go across, we go up that little distance there
will be 2𝒙
for whatever 𝒙 may be
which means - let's go for blue -
in there
we have some angle πœƒ
now actually, we have got two straight lines here
we've got one straight line of the beam as if it carried straight on
the other straight line here which is the tangent
so we know that opposite angle over there
that must also be πœƒ
and that that angle in equals the angle out, relative to the mirror
so when it bounces off
the parabola there
this angle it makes with the tangent is also πœƒ
so in fact, it exits on an overall angle so that entire angle all the way around there
is simply twice whatever the angle of the tangent is at the point where it impacts
so where has that left us?
Well, what we really care about is the focal point
somewhere over here
when that beam carries on
it is going to hit right there
so let's call that point β„Ž
and what do we know? Well, we know where it started.
we know that when the beam hits, at point 𝒙
if you were to carry that across, by definition, it's a parabola
that point on the vertical axis is going to be the point 𝒙²
so we need to know
how much lower
than 𝒙² is it going to be
how far is it going to drop
or what is it's distance 𝑑
now bear in mind the drop is not a great way to put that because there are some points where the light would come in
here and then bounce back up
to the focal point, so it's not necessarily dropping
but I'm going to use 𝑑 as a handy way to do distance
and I find that dropping is an easy way to imagine it
to get that 𝑑, we have to start with the gradient of our tangent, and we already know because it's the derivative
that the gradient, I'm going to use a triangle for that, and a subscript T
equals 2𝒙
What we care about now is what's the new gradient of the beam flying off there
so we want to find what is the gradient again B for the beam.
To do that, we need to see what doubling the angle does to the gradient.
because we've gone from one angle from vertical
to the initial tangent
and then we've doubled that angle out to where the beam is
and for this, we're going to need a bit of trigonometry
ok, so originally we had that the tan (opposite / adjacent) 2𝒙
the tangent of the angle is 1/gradient
in a situation like this
we're now going from tan of a single πœƒ
to tan of double that angle, 2πœƒ
and thankfully, there are a whole lot of double angle formulae
you just look up for any trigonometric function
if you double it's angle what is the result
and of course, you could derive these yourself
I highly recommend it
it's good fun
but in this case,
we're just going to look up
that if you double the angle
of tan
you end up with twice whatever that tan of πœƒ was
divided by 1-tan(πœƒ), all squared
and when we square a trig function we don't put the squared out here
because it looks like we might be squaring the angle
we hide it in there
it's a bit ridiculous
but it helps avoiding mistakes
okay, now we just plug this in
so that is going to equal
two lots of 1/2𝒙, so 2/2𝒙
under that we have 1 minus the square of that, so
1 squared is still 1, and 2 squared is 4𝒙²
to tidy this up, I'm going to multiply the top and the bottom by 4𝒙²
so I'm effectively multiplying it by 1
I'm not going to change it
but that means that that cancels out, that goes there
4𝒙, that becomes 1
that's going to become 4𝒙²
4𝒙²-1
but that is our tan, that is our opposite over our adjacent
to get it back to just be the gradient, we simply flip it the other way up
so the gradient
of the beam
is definitely going to be
4𝒙²-1/4𝒙
for those of you not used to using chalkboards,
I apologise for the sounds that the chalk makes,
it must sound horrific to your delicate whiteboard ears
Let's keep this derivation party going on the other side
so now we know
when the beam comes down
to any point 𝒙
and it hits the parabola,
it gets bounced off
it's new gradient is that
so that for every one step back
towards the vertical axis, it takes, it drops down by that much
and how many steps does it go? Well it goes a distance of 𝒙
and so we can work out
that drop, or 𝑑 distance there
is the gradient
times the distance it's travelled
which is 𝒙, there's going to be some fantastic cancelling out going on there
it's simply going to equal 𝒙²-ΒΌ
and finally, we're going to bring the whole thing home, what is that height there?
how far above the origin?
is our focal point? for any landing point 𝒙
well, it's 𝒙² minus the drop
so that height is going to equal 𝒙² minus the drop of 𝒙²-ΒΌ
Hang on a second!
the drop is exactly one quarter of a unit less than 𝒙²
it's less than that whole distance there
this all cancels out
the height above the origin equals exactly one quarter.
the truly marvellous thing about our final result
is even though we started at some generic point 𝒙
that could have been anywhere on the horizonal axis
it always hits β„Ž on the vertical axis
and there are no 𝒙s left in that result
that is independent of 𝒙
it doesn't matter where the beam starts
it could be over here, it could be over there
somewhere
all beams of light
parallel to the vertical axis
will all converge
and hit the same focal point right in the centre
and yes, we have just done this for the very simple 𝑦=𝒙² parabola
we could do it for a far more general one
and yes, the working out would be a lot more complicated
therefore, fun,
but we actually don't have to do it
it turns out
if you do it for one parabola,
you've actually done it for all of them
but that is a different story, for a different video.
wink wink, coming soon, watch this space.
this truly is the end of the video
this time. Normally, I'd say something like subscribe to my channel
but franky, if you've watched this far though this video,
and the entire proof that I came up with
then you must already be subscribed
you require no further convincing
just thank you so much for watching me do a mass proof about parabolas