P
prime
TID Board Of Directors
- Dec 31, 2011
- 1,178
- 254
So I was thinking about a problem that some people may have run into in the past. Suppose you made two batches of whatever but one batch was stronger than the other. For instance, you made a 250mg/ml batch and then in another batch you ****ed up and the result was a 175mg/ml batch. You don't like the 250ml batch because it's to thick and the 175mg batch dose is just weird and awkward. You decide that 200mg/ml is what you want because it's a nice round number and it's a standard dose per ml amongst the pharmacies and manufacturers.
In hopes of making this simple I wondered if someone on this board had a formula to make this a reality. Then I thought that it shouldn't be that difficult to quickly come up with something. I thought it could be as simple as using a single variable such as x for amount used and apply it to both the high solution and the low solution but that path proved to be more challenging for my analyticaly challenged brain to master. So I settled on using 3 variables and having the result as 1ml broken down in percentage of the two solutions resulting in the exact strength as requested.
So obviously mixing the two in equal amounts won't get you to 200mg/ml. It will get you to 212.5mg/ml. So you say well that's close enough and case closed for you. Well, this thread isn't meant to be dirty and close. It was made to be right on the mark. Exactomundo. Precise. Etc. So 200mg/ml will be the required dose in this prescribed reality.
So I played around with a pencil and a few flip cards and below is the convoluted formula I came up with:
b[(a-x)/(x-b)]/[(a-x)/((x-b)+1)]
+a(1-(((a-x)/(x-b))/((a-x)/(x-b)+1))
=final solution in x strength in 1ml
where
Starting high solution per ml = a (example 250mg/ml)
Starting low solution per ml = b (example 175mg/ml)
goal solution = x (example 200mg/ml)
Please if anybody can reduce this further by all means please contribute.
So it looks like this with the variables replaced with the examples.
175*((250-200)/(200-175))/(250-200)/((200-175)+1)+
250*(1-(250-200)/(200-175))/(250-200)/((200-175)+1))
=200mg/ml
further reduced to the following:
175*((50/25))/(50/25+1)+
250*(1-((50/25))/(50/25+1))
=200mg/ml
reduced again:
175(2/3)+
250(1-2/3)=
200mg/ml
So let's review.
175*2/3ml is 116.66667mg
250*1/3ml is 83.33333mg
add together and we arrive at 200mg/ml If you need 10mls total you just multiply each solution amounts by 10. Simple to extrapolate to larger volumes.
This formula will work for any desired value in between the a and b concentration values. And also any value you assign to a and b. Example, a is 362mg/ml and b is 167mg/ml it will still work So if you wanted to have 217ml it will still work but the fractions won't be as simple as the example above but resulting answer will be right on the mark at x value.
I hope you find this little math problem helpful in solving your future dilution dilemas.
In hopes of making this simple I wondered if someone on this board had a formula to make this a reality. Then I thought that it shouldn't be that difficult to quickly come up with something. I thought it could be as simple as using a single variable such as x for amount used and apply it to both the high solution and the low solution but that path proved to be more challenging for my analyticaly challenged brain to master. So I settled on using 3 variables and having the result as 1ml broken down in percentage of the two solutions resulting in the exact strength as requested.
So obviously mixing the two in equal amounts won't get you to 200mg/ml. It will get you to 212.5mg/ml. So you say well that's close enough and case closed for you. Well, this thread isn't meant to be dirty and close. It was made to be right on the mark. Exactomundo. Precise. Etc. So 200mg/ml will be the required dose in this prescribed reality.
So I played around with a pencil and a few flip cards and below is the convoluted formula I came up with:
b[(a-x)/(x-b)]/[(a-x)/((x-b)+1)]
+a(1-(((a-x)/(x-b))/((a-x)/(x-b)+1))
=final solution in x strength in 1ml
where
Starting high solution per ml = a (example 250mg/ml)
Starting low solution per ml = b (example 175mg/ml)
goal solution = x (example 200mg/ml)
Please if anybody can reduce this further by all means please contribute.
So it looks like this with the variables replaced with the examples.
175*((250-200)/(200-175))/(250-200)/((200-175)+1)+
250*(1-(250-200)/(200-175))/(250-200)/((200-175)+1))
=200mg/ml
further reduced to the following:
175*((50/25))/(50/25+1)+
250*(1-((50/25))/(50/25+1))
=200mg/ml
reduced again:
175(2/3)+
250(1-2/3)=
200mg/ml
So let's review.
175*2/3ml is 116.66667mg
250*1/3ml is 83.33333mg
add together and we arrive at 200mg/ml If you need 10mls total you just multiply each solution amounts by 10. Simple to extrapolate to larger volumes.
This formula will work for any desired value in between the a and b concentration values. And also any value you assign to a and b. Example, a is 362mg/ml and b is 167mg/ml it will still work So if you wanted to have 217ml it will still work but the fractions won't be as simple as the example above but resulting answer will be right on the mark at x value.
I hope you find this little math problem helpful in solving your future dilution dilemas.
Last edited:
if someone has a simpler approach it would be welcome to see. It resulted in a formula I thought would be interesting to share. 
