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Topic: Colloidal Particles on the Head of a Pin  (Read 8908 times)

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Offline Gosseyn

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Colloidal Particles on the Head of a Pin
« on: March 12, 2009, 04:28:11 PM »
The head of a pin 1.5mm in diameter has a diameter of 1500 microns.  Pi R-squared makes the area of the head of the pin equal to 1767145.867644 microns.

The largest colloidal particle would be about .01 microns in diameter. Pi R-squared is pi times .005 or .015707996326795, the area that the particle would occupy.  Thus the number of colloidal particles .01 microns in diameter that can fit on the head of a pin that size would roughly be 1767145.867644 divided by .015707996326795, or about 112,500,000.

Obviously that is very rough because the particles aren't perfect circles, but how's my math on that?   

Offline macman104

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Re: Colloidal Particles on the Head of a Pin
« Reply #1 on: March 12, 2009, 04:55:15 PM »
The head of a pin 1.5mm in diameter has a diameter of 1500 microns.  Pi R-squared makes the area of the head of the pin equal to 1767145.867644 microns.
That looks good
Quote
The largest colloidal particle would be about .01 microns in diameter. Pi R-squared is pi times .005 or .015707996326795, the area that the particle would occupy.
Check the math here.  You didn't do r^2, just r.

Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #2 on: March 12, 2009, 05:00:17 PM »
The head of a pin 1.5mm in diameter has a diameter of 1500 microns.  Pi R-squared makes the area of the head of the pin equal to 1767145.867644 microns.

The largest colloidal particle would be about .01 microns in diameter. Pi R-squared is pi times .005 or .015707996326795, the area that the particle would occupy.  Thus the number of colloidal particles .01 microns in diameter that can fit on the head of a pin that size would roughly be 1767145.867644 divided by .015707996326795, or about 112,500,000.

Obviously that is very rough because the particles aren't perfect circles, but how's my math on that?   


Oops, I see that is wrong already.  I'll try again later.

Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #3 on: March 12, 2009, 05:03:01 PM »
The head of a pin 1.5mm in diameter has a diameter of 1500 microns.  Pi R-squared makes the area of the head of the pin equal to 1767145.867644 microns.
That looks good
Quote
The largest colloidal particle would be about .01 microns in diameter. Pi R-squared is pi times .005 or .015707996326795, the area that the particle would occupy.
Check the math here.  You didn't do r^2, just r.

I must have caught that about the same time as you did, since I didn't see your post before now.

Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #4 on: March 12, 2009, 05:18:26 PM »
The head of a pin 1.5mm in diameter has a diameter of 1500 microns.  Pi R-squared makes the area of the head of the pin equal to 1767145.867644 microns.
That looks good
Quote
The largest colloidal particle would be about .01 microns in diameter. Pi R-squared is pi times .005 or .015707996326795, the area that the particle would occupy.
Check the math here.  You didn't do r^2, just r.

Thanks, here is the revision:

The head of a pin 1.5mm in diameter has a diameter of 1500 microns.  PiR2 makes the area of the head of the pin equal to 1767145.867644 microns in area.

The largest colloidal particle would be about .01 microns in diameter. PiR2 is pi times .005 squared or 7.853981633974483e-5, the area of the particle.  Thus the number of colloidal particles .01 microns in diameter that can fit on the head of the pin would be 1767145.867644 divided by 7.853981633974483e-5, or about 22.5 billion. 

How is the math now, and does that make sense (sort of)?       

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Re: Colloidal Particles on the Head of a Pin
« Reply #5 on: March 12, 2009, 06:08:18 PM »
about 22.5 billion.

(1.5e-3/0.01e-6)2, seems OK.

Although if you assume them to be round in shape, they cover only 90.7% of the surface... (π sqrt(3)/6 - unless I did something wrong).
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Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #6 on: March 12, 2009, 07:08:36 PM »
about 22.5 billion.

(1.5e-3/0.01e-6)2, seems OK.

Although if you assume them to be round in shape, they cover only 90.7% of the surface... (π sqrt(3)/6 - unless I did something wrong).

Yes, is that because of the spaces created between touching circles because of their shape?  I was wondering about that. 

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Re: Colloidal Particles on the Head of a Pin
« Reply #7 on: March 12, 2009, 07:17:27 PM »
Yes. But it is a simple geometry. Imagine three touching circles. Their centers form an equilateral triangle. Identical ttriangles cover whole surface, so if you can calculate what part of the triangle is ocupied by the circles, you are there. Simple geometry.

Hint: what part of the full circle you will get from three 60 degree circular sectors?
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Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #8 on: March 14, 2009, 04:06:56 PM »
Yes. But it is a simple geometry. Imagine three touching circles. Their centers form an equilateral triangle. Identical ttriangles cover whole surface, so if you can calculate what part of the triangle is ocupied by the circles, you are there. Simple geometry.

Hint: what part of the full circle you will get from three 60 degree circular sectors?

Well, I don't know simple geometry very well but I used the pythagorean theorm to get the height of the right-angle triangle, then calculated the area of the equilateral triangle and subtracted half the area of the circle and got a figure of about .0000041 as the missing area for each equilateral triangle.  But even if that is right (and it likely isn't, since I struggled with it), how do I know how many equilateral triangles would be on the pin as a whole given that the triangles share some of the same circles?   

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Re: Colloidal Particles on the Head of a Pin
« Reply #9 on: March 14, 2009, 05:24:53 PM »
Well, I don't know simple geometry very well but I used the pythagorean theorm to get the height of the right-angle triangle, then calculated the area of the equilateral triangle and subtracted half the area of the circle and got a figure of about .0000041 as the missing area for each equilateral triangle.

No, that's wrong, you must have miscalculate triangle height.

Quote
But even if that is right (and it likely isn't, since I struggled with it), how do I know how many equilateral triangles would be on the pin as a whole given that the triangles share some of the same circles?   

You don't have to know number of triangles. Circles are not shared - I mean, they are, but the part INSIDE of the triangle is not, and you used only that part in your calculations. So this triangle, repetaed many times, covers whole surface.

Well, not whole. There will be problems at borders. But they don't matter for a pin large enough ;)
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Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #10 on: March 14, 2009, 05:36:00 PM »
Well, I don't know simple geometry very well but I used the pythagorean theorm to get the height of the right-angle triangle, then calculated the area of the equilateral triangle and subtracted half the area of the circle and got a figure of about .0000041 as the missing area for each equilateral triangle.

No, that's wrong, you must have miscalculate triangle height.

Quote
But even if that is right (and it likely isn't, since I struggled with it), how do I know how many equilateral triangles would be on the pin as a whole given that the triangles share some of the same circles?   

You don't have to know number of triangles. Circles are not shared - I mean, they are, but the part INSIDE of the triangle is not, and you used only that part in your calculations. So this triangle, repetaed many times, covers whole surface.

Well, not whole. There will be problems at borders. But they don't matter for a pin large enough ;)

OK, so I would just divide the area of the pin by the area of the equilateral triange and then multiply that by the missing area?  Since I got it wrong, I'll try to work the problem of the area of the triangle again.

Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #11 on: March 14, 2009, 06:30:26 PM »
I'm still getting the height of the triangle as 0.0086.  Is that correct?

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Re: Colloidal Particles on the Head of a Pin
« Reply #12 on: March 14, 2009, 06:39:27 PM »
Sorry, could be I have misunderstood your post. You have not listed units so I assumed you did the most obvious thing - you calculated area of a triangle with side of the length 1.

OK, so I would just divide the area of the pin by the area of the equilateral triange and then multiply that by the missing area?

You can do it do this way, but there is no need for that. Not that units cancel out, so you don't have to work with a real size of the molecule. Regardless of the molecule size ratio of the area occupied by molecules to the total surface is identical, and can be easily found assuming molecule radius to be 1. That'll give an equilateral triangle with a=1.
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Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #13 on: March 14, 2009, 08:53:52 PM »
Sorry, could be I have misunderstood your post. You have not listed units so I assumed you did the most obvious thing - you calculated area of a triangle with side of the length 1.

OK, so I would just divide the area of the pin by the area of the equilateral triange and then multiply that by the missing area?

You can do it do this way, but there is no need for that. Not that units cancel out, so you don't have to work with a real size of the molecule. Regardless of the molecule size ratio of the area occupied by molecules to the total surface is identical, and can be easily found assuming molecule radius to be 1. That'll give an equilateral triangle with a=1.

I don't understand.  If the molecule has a radius of 1, wouldn't the side of the triangle be 2?  Then the area of the triangle would be 1.73.

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Re: Colloidal Particles on the Head of a Pin
« Reply #14 on: March 15, 2009, 04:35:49 AM »
Sorry, it would be 2, my mistake. I was thinking about diameter, not about radius.

And 1.73 is just a square root of 3, isn't it? So for side 2 that's correct.

http://en.wikipedia.org/wiki/Equilateral_triangle
« Last Edit: March 15, 2009, 04:47:31 AM by Borek »
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