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 The time now is Sun Dec 08, 2013 10:09 pm All times are UTC - 6 View posts in this forum since last visit View unanswered posts in this forum Calendar
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norman182
Boot

Joined: 16 Apr 2006
Posts: 14

#205 Whipsmart Hot Fudge

our famous HOT FUDGE CONE
..take the crispiest, crunchiest cone
..scoop in a big, 2 inch diameter ball of vanilla ice-cream
..add four identical scoops of chocolate icecream inside the cone, making sure they're arranged so as to all touch the top ot the vanilla
..pour over lashings of hot fudge sauce
..cover in whipped cream
..put a cherry on the top.

At whipsmart, we've been making our famous hot fudge cone for over 50 years now and it's still one of our biggest sellers. that's because we like
to make sure our customers get the most ice cream possible

Unfortunately we forgot to include a detail on our new instruction poster - how much chocolate ice cream! given that our cones have a 30 degree apex, what's the biggest diameter scoops of chocolate ice cream we can use in inches
( to 2 decimal places)?

Posted: Sun Jul 30, 2006 9:21 am
Sammykat
Boot

Joined: 03 Jul 2006
Posts: 40

My maths is terrible so this may well be wrong..

 Spoiler (Rollover to View): Circumference of the vanilla ball is 2inches, meaning the cone has a circumferance of at least that. Pi = 3.142, so I changed the inches to cm (15), then changed it back, so its 5.90 inches round. That divided by 4 = 1.475inches, so in theory you could plop them on top of the vanilla. I don't think we need to know how much more cone there is above that, because regardless of the extra circumference the chocolate can't all touch the vanilla if its bigger. (Well, it could, in the real world, where ice cream melts.)

Without finding a protractor, (may be hard as no-one in my house has been at school for a few years) this is the best I can do. Feel free to throw my maths out the window though.

Posted: Mon Jul 31, 2006 6:33 am
Curlytek
Veteran

Joined: 30 Jul 2005
Posts: 112
Location: Melbourne, Australia

Hi Sammykat, welcome to the world of PXC!

The vanilla ball has a diammeter of 2 inches, not a circumference (according to the above transcription), which is the longest straight line through the centre of the sphere, between two points of the spherical surface (ie. 2 * the radius).

 Spoiler (Rollover to View): I also think that the packing of spheres is a bit more complex than you have described, if you think of the problem from above you need to have four circles of equal diammeter (the chocolate scoops) fitting within a bigger circle (the cone). To get the answer, you need to work out at what height of the cone the centre of the chocolate balls (hmmm, sounds a bit like a South Park reference) will fit, given that they all must touch the vanilla ball. It may even be that the chocolate scoops don't even touch each other, but I will need to do more drawing/thinking/maths to work this one out.

Posted: Mon Jul 31, 2006 8:44 pm
fretty
Decorated

Joined: 19 Nov 2004
Posts: 281
Location: South Yorkshire, England

But for the maximum diameter, isn't that when they all do touch?

Posted: Tue Aug 01, 2006 2:38 am
Sammykat
Boot

Joined: 03 Jul 2006
Posts: 40

Oops, I didnt make it very clear. I messed up the wording on my first sentence. Circumferunce = diameter. It's cos I was thinking of the working as opposed to putting it into english.

 Spoiler (Rollover to View): The Vanilla Circumference is 5.90, so the cone has to be that at a certain point. I don't know how much of an increase in diameter the section above the vanilla will have (as I can't find anything to draw it out) - but the max size is limited by the diameter (after working the circumferance then x 4) equalling that circ. I have a feeling my workings utterly wrong anyway, I took the circumference and divided it by 4 which give the circum of the chocolate, not the diameter. Just realised I didn't use it to work the diameter. Which is, after some rummaging to remember how to do it, 0.4695(inches) I doubt it'll be anymore then 0.5?

Hope this gives someone with better maths then me a good starting point ^_^; And this, ladies and gents, is one of the many reasons I failed maths.

Posted: Tue Aug 01, 2006 3:55 am
Curlytek
Veteran

Joined: 30 Jul 2005
Posts: 112
Location: Melbourne, Australia

 fretty wrote: But for the maximum diameter, isn't that when they all do touch?

My thinking there fretty was that since all four chocolate scoops have to touch the vanilla, and because of gravity also need to touch the cone, then their biggest size may not necessarily allow them to touch each other. But you are probably right, now that I have thought about it further...

Posted: Tue Aug 01, 2006 6:20 am
Generica
Greenhorn

Joined: 01 Aug 2006
Posts: 5
Location: Leeds, UK

 Spoiler (Rollover to View): I didn't use the circumference to solve the card. Try drawing a detailed diagram of the top and side views of the cone with the five scoops of ice cream inside. From there, it's a matter of applying lots of trig to reach a solution. I'm trying really hard to think of a good hint that doesn't just give the card away, as I don't want to spoil everyone's fun. However, this card doesn't really lend itself to nifty hints.

Posted: Tue Aug 01, 2006 4:01 pm
Muffin
Unfettered

Joined: 30 Oct 2003
Posts: 306
Location: UK, Leicester

To be honest, I think the easiest way to solve this one is with a large piece of graph paper, a compass and a protractor. Although saying that, you do need to give you answer to the closest 0.254 mm so maybe that won't be accurate enough. I'm sure there is a trig way to do this, but it will be darn complicated!

Just one note though, for the vanilla scoop - it won't touch the sides at its max. diameter - this would require the cone to be a cylinder, which wouldn't be great for eating your ice cream out of. This is where the trig comes in!
_________________
100 of 333

Posted: Wed Aug 02, 2006 2:53 am
Sammykat
Boot

Joined: 03 Jul 2006
Posts: 40

D: Dammit.

I'm gonna have to hope someone can tell me the answer then. I really am that bad at maths. I dont even know how to do trig, at all. I can't even remember studying it.

Posted: Wed Aug 02, 2006 5:47 am
almagest
Boot

Joined: 11 Jul 2006
Posts: 45
Location: London

Unless I am missing some elegant solution, this is a messy geometry problem. Note first the slight ambiguity in the meaning of apex angle. It means that the angle between the axis of symmetry and a line in the surface through the apex is 15 deg.
If you look at the four choc spheres (each radius x), then the distance of their centres from the axis of symmetry is x root 2. (Just take a section through all their centres). Now take a section through the vanilla centre, the apex and two of the choc centres. You know the angles and can easily get an equation for x.

Posted: Wed Aug 02, 2006 3:30 pm
anansi
Boot

Joined: 11 Mar 2006
Posts: 51
Location: Leeds, UK

almagest, are you meaning

 Spoiler (Rollover to View): that you work out x from a triangle with two (opposite) chocolate centres and the vanilla centre as its vertices, or a triangle with two (opposite) chocolate centres and the point of the cone as its vertices (so that the centre of the vanilla scoop lies on the plane formed by the triange? Either way, I'm struggling to find the angles in the riangle. Am I missing some really simple way to calculate it? I should really get this, my highest mark at uni was in Geometry!!

Jack

Posted: Thu Aug 03, 2006 6:57 am
almagest
Boot

Joined: 11 Jul 2006
Posts: 45
Location: London

Well, I said it was messy! I used more than one triangle.

Posted: Thu Aug 03, 2006 7:54 am
fretty
Decorated

Joined: 19 Nov 2004
Posts: 281
Location: South Yorkshire, England

Can someone check if this is correct:
 Spoiler (Rollover to View): Calling x the radius Making a triangle from above in a horizontal plane with hypotenuse 2x (from two adjoining chocolate spheres), the distance between the centre of the cone and the centre of one of the chocolate spheres can be worked out by using pythagoras to be x root 2 as almagest said. Then by making a triangle with angle of 15 degrees, an opposite side of length x root 2, and a hypotenuse of length x+1 (as the two radii combine), you can say that: sin 15 = (x root 2)/(x+1) Solving gives x=0.224... or 0.22 (to 2 d.p.)

EDIT: changed the end value

Posted: Thu Aug 03, 2006 7:57 am
Last edited by fretty on Thu Aug 03, 2006 8:22 am; edited 1 time in total
Sophiecat
Decorated

Joined: 08 Apr 2006
Posts: 171
Location: North East UK

I got the same equation, but when I solve it I get
 Spoiler (Rollover to View): 0.22" to 2d.p.

Unfortunately I'm locked out until 10pm (not helped by trying my first two solves in cm!)so can't check either.

Posted: Thu Aug 03, 2006 8:16 am
fretty
Decorated

Joined: 19 Nov 2004
Posts: 281
Location: South Yorkshire, England

Sorry, yeah that's what I get now, bloody google.

Posted: Thu Aug 03, 2006 8:21 am
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