Differences between peak dose and peak dose rate

  • #1

Summary:

Is peak dose and peak dose rate the same thing or different?
What is the difference between peak dose 100 kRads (Si) and peak dose rate 10 x 10^12 Rads (Si)/s? Can someone explain?
 

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  • #2
Vanadium 50
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One is total and the other is per unit time. Like work and time. Or pay per hour and the size of your paycheck.
 
  • #3
One is total and the other is per unit time. Like work and time. Or pay per hour and the size of your paycheck.
So, if I understand it correctly, peak dose rate of 10 x 10^12 Rads (Si)/s means 10,000,000,000,000 Rads per second? Isn't it an exaggeration?
 
  • #4
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A reference of where you saw these numbers would help.
 
  • #6
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Ok. And what is the problem? The peak dose rates referenced in the article are for a FWHM 20ns pulse. Because of the short pulse, the dose (which is the integrated dose -> area under the peak dose curve) is substantially less.
 
  • #7
Ok. And what is the problem? The peak dose rates referenced in the article are for a FWHM 20ns pulse. Because of the short pulse, the dose (which is the integrated dose -> area under the peak dose curve) is substantially less.
So, the peak dose actually is not 10,000,000,000,000 Rads like what we see in Figure 5? The part which I don't understand is why did they put times 10^12 Rads/s in the figure if it is actually less.
 
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  • #8
Vanadium 50
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First, your obsession with Hermes-III is really not helpful. You seem to have decided that it is uniquely dangerous first and then started looking for facts to support this pre-conclusion.

Second, the difference between peak and total is not unique to Hermes, nor accelerators, nor even physics.
Suppose you get paid $100 a day. Now suppose you were paid $100 for an hour, but had to take the rest of the day off. Now suppose you were paid $100 for an second, but had to take the rest of the day off. Now suppose you were paid $100 for an microsecond, but had to take the rest of the day off. In which case do you make the most money?

In any of the circumstances does the pay rate of $3,1000,000,000,000/year make sense?
 
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  • #10
First, your obsession with Hermes-III is really not helpful. You seem to have decided that it is uniquely dangerous first and then started looking for facts to support this pre-conclusion.

Second, the difference between peak and total is not unique to Hermes, nor accelerators, nor even physics.
Suppose you get paid $100 a day. Now suppose you were paid $100 for an hour, but had to take the rest of the day off. Now suppose you were paid $100 for an second, but had to take the rest of the day off. Now suppose you were paid $100 for an microsecond, but had to take the rest of the day off. In which case do you make the most money?

In any of the circumstances does the pay rate of $3,1000,000,000,000/year make sense?
I tried to find the specs of other accelerators but so far Hermes-III have the most detailed specs.

I don't know if my calculation is correct. But one second is equal to 1000000000 nanosecond. Let's assume the peak dose rate is 15 x 10^12 Rads per second. 15 x 10^12/s is 15000000000000 per one second. I divide 15000000000000 by 1000000000 nanosecond and I get 15000. If we reduced it by 20 nanosecond, then I get 30000 Rads. So, in 20 nanosecond the total dose is 30000 Rads. Is that correct?
 
  • #11
jbriggs444
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I don't know if my calculation is correct. But one second is equal to 1000000000 nanosecond. Let's assume the peak dose rate is 15 x 10^12 Rads per second. 15 x 10^12/s is 15000000000000 per one second. I divide 15000000000000 by 1000000000 nanosecond and I get 15000.
You divide 15 x 10^12 rads per second by 1000000000 nanoseconds per second and get 15000 rads per nanosecond. Yes, that would be a correct calculation.
If we reduced it by 20 nanosecond, then I get 30000 Rads. So, in 20 nanosecond the total dose is 30000 Rads. Is that correct?
15000 rads per nanosecond with an exposure of 20 nanoseconds gives 300000 rads.

You lost me on the "reduced it by 20 nanosecond" bit. What are we reducing? And why are we left with 20 nanoseconds?
 
  • #12
You divide 15 x 10^12 rads per second by 1000000000 nanoseconds per second and get 15000 rads per nanosecond. Yes, that would be a correct calculation.

15000 rads per nanosecond with an exposure of 20 nanoseconds gives 300000 rads.

You lost me on the "reduced it by 20 nanosecond" bit. What are we reducing? And why are we left with 20 nanoseconds?
Because the pulsewidth is just 20 nanosecond. The peak dose (not peak dose rate) is around >100 kRads according to the specs of the accelerator, which is lower than 300000. So, I think 300000 will be the absolute max.
 
  • #13
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Because the pulsewidth is just 20 nanosecond. The peak dose (not peak dose rate) is around >100 kRads according to the specs of the accelerator, which is lower than 300000. So, I think 300000 will be the absolute max.
So we need to reconcile a calculation that multiplies the peak dose rate by the pulse width and gets a figure that is three times as high as the peak dose.

Do you know about integrals?
 
  • #14
So we need to reconcile a calculation that multiplies the peak dose rate by the pulse width and gets a figure that is three times as high as the peak dose.

Do you know about integrals?
I have to tell you, my mathematics skills are bad. :frown:

I asked one of my friend and he responded, the difference between peak dose and peak dose rate indicates the pulsewidth is not square. If the pulse is square, dose rate will be total dose divided by the length of the pulse in seconds.
 
  • #15
jbriggs444
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I have to tell you, my mathematics skills are bad. :frown:

I asked one of my friend and he responded, the difference between peak dose and peak dose rate indicates the pulsewidth is not square. If the pulse is square, dose rate will be total dose divided by the length of the pulse in seconds.
Yes. That is the 'gist of what I was going to be saying.

If you graph out the dosage rate over time and look at the area under the curve, you'll get the total dose. For a square wave form, the area is pulse width times pulse height just as one would expect. But for a non-square wave, the area will be something less than the peak height times the total width.

In mathematics, an "integral" is the area under a curve.
 
  • #16
Yes. That is the 'gist of what I was going to be saying.

If you graph out the dosage rate over time and look at the area under the curve, you'll get the total dose. For a square wave form, the area is pulse width times pulse height just as one would expect. But for a non-square wave, the area will be something less than the peak height times the total width.

In mathematics, an "integral" is the area under a curve.
I think I finally get it, the peak dose rate is the total dose divided by the length of the pulse that is converted into seconds. The peak dose rate which is 5 x 10^12 is equal to 5000000000000. I divided the peak dose which is 100000 by 0.00000002 second which equals to 20 nanosecond and I got 5000000000000 which is the same when we calculate the peak dose rate. So, the 5 x 10^12 basically means 100kRads. This is confusing for someone who is not really good in mathematics like me.
 

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