Calculate gradient %A, %B total volumes

[Bill]

Hi Group,

Trying to figure out how to calculate the total volume of A and B used
in a linear binary gradient. For example, If I start with 95%A and
ramp to 95 % B in 1 min, what would be the total volume of A and B is
used (assume 1 mL/min)? I can calculate the slope and intercept easy
enough (2 equations, 2 unknowns), but not sure how to get the total
volume used with it. With this example, I would guess it would be the
intersection midpoint or 50% A, 50%B times the flow rate, i.e., 0.5 mL
of A and 0.5 mL B

Not sure if the same would be true if I start with 50:50 A:B and ramp
to 20:80 A:B in say 10 min. At the 5 min. midpoint, it would be 35:65
so if it is a linear gradient would it be %A or B times flow rate
accounting for initial offset of 50:50.

I think we did this years ago as a deff. eq. or calculus class
problem, but it's been years. My particular gradient has different
steps and holds so the isochratic holds are easy to add to the
segment, but the linear gradients have me thinking. Maybe (Iso flow 1
+ gradient midpoint 1 + iso hold 2 +gradient midpoint 2...). I'll
eventually do this experimentally but don't want to stop my system and
get a good initial A and B volume! Any help greatly appreciated!

Best Regards,

Bill

[Chip]

You need to take the area under the curve where you plot ml/min for A
and B vs. time. Assume the simplest case: a linear gradient from 0%A
to 100%B in 10 minutes. (Yes, I know that's the wrong convention for
a reversed-phase gradient.) Assume a 2 ml/min constant flow rate and
ignore nonideal solubility (e.g. methanol and water). At time 0 you
have 0 ml/min A and at time 10 you have 2 ml A/min. The area under
the curve is a simple triangular area A = base*height/2. In this
case, you would use 10 ml of A. If you had used a constant 100% A
gradient, you would have used 20 ml.


If you have a more complex gradient, you can break it up into segments
and sum the area of each segment. If you don't have a linear
gradient, use numerical integration (e.g. the trapezoidal rule).


Chip

Post by Bill
Hi Group,
Trying to figure out how to calculate the total volume of A and B used
in a linear binary gradient.  For example, If I start with 95%A and
ramp to 95 % B in 1 min, what would be the total volume of A and B is
used (assume 1 mL/min)?  I can calculate the slope and intercept easy
enough (2 equations, 2 unknowns), but not sure how to get the total
volume used with it.  With this example, I would guess it would be the
intersection midpoint or 50% A, 50%B times the flow rate, i.e., 0.5 mL
of A and 0.5 mL B
Not sure if the same would be true if I start with 50:50 A:B and ramp
to 20:80 A:B in say 10 min.  At the 5 min. midpoint, it would be 35:65
so if it is a linear gradient would it be %A or B times flow rate
accounting for initial offset of 50:50.
I think we did this years ago as a deff. eq. or calculus class
problem, but it's been years.  My particular gradient has different
steps and holds so the isochratic holds are easy to add to the
segment, but the linear gradients have me thinking.  Maybe (Iso flow 1
+ gradient midpoint 1 + iso hold 2 +gradient midpoint 2...).   I'll
eventually do this experimentally but don't want to stop my system and
get a good initial A and B volume!  Any help greatly appreciated!
Best Regards,
Bill

[Chip]

I'll blame it on jet lag -- 0%A to 100%B at 10 minutes is normal for a
reverse-phase gradient.

Post by Chip
You need to take the area under the curve where you plot ml/min for A
and B vs. time.  Assume the simplest case: a linear gradient from 0%A
to 100%B in 10 minutes.  (Yes, I know that's the wrong convention for
a reversed-phase gradient.)  Assume a 2 ml/min constant flow rate and
ignore nonideal solubility (e.g. methanol and water).  At time 0 you
have 0 ml/min A and at time 10 you have 2 ml A/min.   The area under
the curve is a simple triangular area A = base*height/2.  In  this
case, you would use 10 ml of A.   If you had used a constant 100% A
gradient, you would have used 20 ml.
If you have a more complex gradient, you can break it up into segments
and sum the area of each segment.   If you don't have a linear
gradient, use numerical integration (e.g. the trapezoidal rule).
    Chip

Post by Bill
Hi Group,
Trying to figure out how to calculate the total volume of A and B used
in a linear binary gradient.  For example, If I start with 95%A and
ramp to 95 % B in 1 min, what would be the total volume of A and B is
used (assume 1 mL/min)?  I can calculate the slope and intercept easy
enough (2 equations, 2 unknowns), but not sure how to get the total
volume used with it.  With this example, I would guess it would be the
intersection midpoint or 50% A, 50%B times the flow rate, i.e., 0.5 mL
of A and 0.5 mL B
Not sure if the same would be true if I start with 50:50 A:B and ramp
to 20:80 A:B in say 10 min.  At the 5 min. midpoint, it would be 35:65
so if it is a linear gradient would it be %A or B times flow rate
accounting for initial offset of 50:50.
I think we did this years ago as a deff. eq. or calculus class
problem, but it's been years.  My particular gradient has different
steps and holds so the isochratic holds are easy to add to the
segment, but the linear gradients have me thinking.  Maybe (Iso flow 1
+ gradient midpoint 1 + iso hold 2 +gradient midpoint 2...).   I'll
eventually do this experimentally but don't want to stop my system and
get a good initial A and B volume!  Any help greatly appreciated!
Best Regards,
Bill

[Bill]

Post by Chip
I'll blame it on jet lag -- 0%A to 100%B at 10 minutes is normal for a
reverse-phase gradient.

Post by Chip
You need to take the area under the curve where you plot ml/min for A
and B vs. time.  Assume the simplest case: a linear gradient from 0%A
to 100%B in 10 minutes.  (Yes, I know that's the wrong convention for
a reversed-phase gradient.)  Assume a 2 ml/min constant flow rate and
ignore nonideal solubility (e.g. methanol and water).  At time 0 you
have 0 ml/min A and at time 10 you have 2 ml A/min.   The area under
the curve is a simple triangular area A = base*height/2.  In  this
case, you would use 10 ml of A.   If you had used a constant 100% A
gradient, you would have used 20 ml.
If you have a more complex gradient, you can break it up into segments
and sum the area of each segment.   If you don't have a linear
gradient, use numerical integration (e.g. the trapezoidal rule).
    Chip

Post by Bill
Hi Group,
Trying to figure out how to calculate the total volume of A and B used
in a linear binary gradient.  For example, If I start with 95%A and
ramp to 95 % B in 1 min, what would be the total volume of A and B is
used (assume 1 mL/min)?  I can calculate the slope and intercept easy
enough (2 equations, 2 unknowns), but not sure how to get the total
volume used with it.  With this example, I would guess it would be the
intersection midpoint or 50% A, 50%B times the flow rate, i.e., 0.5 mL
of A and 0.5 mL B
Not sure if the same would be true if I start with 50:50 A:B and ramp
to 20:80 A:B in say 10 min.  At the 5 min. midpoint, it would be 35:65
so if it is a linear gradient would it be %A or B times flow rate
accounting for initial offset of 50:50.
I think we did this years ago as a deff. eq. or calculus class
problem, but it's been years.  My particular gradient has different
steps and holds so the isochratic holds are easy to add to the
segment, but the linear gradients have me thinking.  Maybe (Iso flow 1
+ gradient midpoint 1 + iso hold 2 +gradient midpoint 2...).   I'll
eventually do this experimentally but don't want to stop my system and
get a good initial A and B volume!  Any help greatly appreciated!
Best Regards,
Bill

Thank you sir! Area under the curve integral... area of a triangle,
makes perfect sense. I doubt I'll ever have to use Simpson's rule for
non-linear gradients, but easy enough to model in Excel. Again, many
thanks!

Best Regards,

Bill

[Chip]

Post by Chip
I'll blame it on jet lag -- 0%A to 100%B at 10 minutes is normal for a
reverse-phase gradient.

Post by Chip
You need to take the area under the curve where you plot ml/min for A
and B vs. time.  Assume the simplest case: a linear gradient from 0%A
to 100%B in 10 minutes.  (Yes, I know that's the wrong convention for
a reversed-phase gradient.)  Assume a 2 ml/min constant flow rate and
ignore nonideal solubility (e.g. methanol and water).  At time 0 you
have 0 ml/min A and at time 10 you have 2 ml A/min.   The area under
the curve is a simple triangular area A = base*height/2.  In  this
case, you would use 10 ml of A.   If you had used a constant 100% A
gradient, you would have used 20 ml.
If you have a more complex gradient, you can break it up into segments
and sum the area of each segment.   If you don't have a linear
gradient, use numerical integration (e.g. the trapezoidal rule).
    Chip

Post by Bill
Hi Group,
Trying to figure out how to calculate the total volume of A and B used
in a linear binary gradient.  For example, If I start with 95%A and
ramp to 95 % B in 1 min, what would be the total volume of A and B is
used (assume 1 mL/min)?  I can calculate the slope and intercept easy
enough (2 equations, 2 unknowns), but not sure how to get the total
volume used with it.  With this example, I would guess it would be the
intersection midpoint or 50% A, 50%B times the flow rate, i.e., 0.5 mL
of A and 0.5 mL B
Not sure if the same would be true if I start with 50:50 A:B and ramp
to 20:80 A:B in say 10 min.  At the 5 min. midpoint, it would be 35:65
so if it is a linear gradient would it be %A or B times flow rate
accounting for initial offset of 50:50.
I think we did this years ago as a deff. eq. or calculus class
problem, but it's been years.  My particular gradient has different
steps and holds so the isochratic holds are easy to add to the
segment, but the linear gradients have me thinking.  Maybe (Iso flow 1
+ gradient midpoint 1 + iso hold 2 +gradient midpoint 2...).   I'll
eventually do this experimentally but don't want to stop my system and

Glad to help. This newsgroup is better than many, but it's always
nice to see on-topic questions!

Post by Chip

Post by Chip

Post by Bill
get a good initial A and B volume!  Any help greatly appreciated!
Best Regards,
Bill

Thank you sir!  Area under the curve integral... area of a triangle,
makes perfect sense.  I doubt I'll ever have to use Simpson's rule for
non-linear gradients, but easy enough to model in Excel.  Again, many
thanks!
Best Regards,
Bill