Dear Mr. Thumbs,
I'm very sorry to hear about the mishap that has so unfairly cut short the life of your greenhouse. I blame only myself. Any glass greenhouse that cannot withstand the impact of a truck traveling at a mere five miles per hour obviously has some serious structural issues. Perhaps I should have told you to lean the glass .05 degrees farther towards the ground…
In any case, I feel terrible, so the least that I can do is to help you with your latest problem. Now I understand that you have a serious pollen problem (somehow on account of my faulty greenhouse design I am sure), and that you have purchased an air purifier in order to solve this problem. From your most recent letter it seems to me that your question is in regards to the time that it will take the air purifier to make your house safely habitable again (something I suppose we all want in a home), which was determined by the health inspector to be when the pollen is at 1/10th of a percent of its present level. If I remember correctly, your air purifier claims to clean the house at a rate of half the air in the house per day, with a pollen removal rate of 100% from the circulated air. You were confused as to why it would not remove all of the pollen from the house in two days, if it cleans half the air in one day.
This is a very legitimate and no doubt common misunderstanding. The reason that the house will take longer than two days to get down to an acceptable air quality is that the circulator functions at a rate of half the house per day. Allow me to clarify. This does not mean that on day one it cleans all of the pollen out of one half, and on day two it cleans all of the pollen out of the other half (figure 1). For this to be true, there would have to be no mixing between the two different sections of air. What is actually happening is that on day one the machine cleans half of the pollen out of the house. Then, before the next day, essentially what happens is that all of the air mixes, and all of the pollen gets spread out throughout the house, so throughout the house there is an even level of pollen, exactly half of what it was the previous day (figure 2). On day two into day three the same thing happens, with the result being air with one quarter of the original air.
This is all based on several assumptions about the state of the house. The first is that there is not net exchange of pollen between the inside of the house and the outside world. Also, the air must be allowed to flow throughout the house, so all areas inside have the same amount of pollen. If these things are true, it is possible, albeit not easy to calculate how long it will take you to reduce your pollen level to 1/10 of a percent of what it currently is.
When I first began to think about this problem, it frustrated me greatly. I could not think of any way to solve it. Then one day I saw something that made everything so obvious and crystal clear. I was walking my dog, and what did I see on the side of the street? A slowly decaying radioactive isotope! "Eureka!", I said aloud, and I smacked myself on the forehead. How could I have not seen it earlier? Isn't it odd that any time you have a problem, no matter what it is related to, the answer is invariably linked to the decay of radioisotopes?
Although I am sure that you understand, perhaps I should explain for any kindergarteners who might have the same question. I won't bore you with the details, but in any case, many compounds have what is called a half life, which says that they will decay, or breakdown, at a rate of half every certain amount of years (my bio book). For instance, if compound x has a half life of 10 years, and starts with 100 units of matter, 10 years later it will have 50 units, and 10 years after that it will have 23 units, and so on.
What I realized was that the pollen in your house is like a radioactive isotope with a half life of one day. Once I knew this, it as easy, because there actually is an established formula for radioactive decay (my calc book). I won't try to explain the details, but I will tell you the formula for your own whimsy (and incase you don't trust me). The formula is
This may seem complicated to you, but in actuality, we already know most of the variables that will be plugged right into the equation. P0 represents the current amount of pollen in the air, which we will give a value of 1000 (because the desired level is 1/1000th of the current level). P represents the desired level of pollen, which we will give a value of 1. k represents the rate of decay per day, which is -.5, and t is the time is will take for P0 to become equal to, or less than P, which we don't know and will be the answer to the question. We start by plugging in the known values,
The next thing that you have to do is to solve for t. To do this we take the natural log of both sides of the equation, and get t by itself.
First we take natural logarithm of both sides:
We then divide both sides of the equation by (-0.5) to get the t all alone:
and finally we blug this into the calculator to get the number of days before your pollen is acceptably low:
By plugging in all of the known values into the equation, I reached a value for t of 13.82 days. You can test this by plugging in all values (including t), except for P, into the equation, and you will get a pollen content of 1, exactly what it needs to be. This means that by the 14th day of air circulation you should be able to live in your house again!
Anyway, I hope that this information was of help to you. I want to thank all of my books and Dusty for the help with the math. And I want to thank Professor Crannell for relaying the message to me.
Good luck with everything,
Andy Margules