FAQ  DD fusion  Energy  Flux  Dose
 Richard Hull
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 Joined: Fri Jun 15, 2001 1:44 pm
 Real name: Richard Hull
FAQ  DD fusion  Energy  Flux  Dose
This effort will discuss only amateur deuteriumdeuterium fusion. It will discuss energy releases, how to figure flux, discuss does rate versus equivalent dose.
All calculations will be carried out to only the nearest tenth part of any value.
DeuteriumDeuterium fusion data
Two reactions occur with about equal probability.
D + D = T (tritium 1.01 MeV) + P (3.02MeV)
D + D = 3He (.82 MeV) + n (2.45 MeV)
NOTE* There is a third reaction possible, but it only occurs once in every ~50,000 DD fusions and that is D+D = 4He + gamma ray.....We will not even consider this reaction in any of our work due to its extreme rarity.
Thus DD fusion is a mixed output of the above two major reactions such that the energy from DD fusion (7.3 MeV) shakes out to the following fractions of the net ongoing reactions.
T – 13.8%
P  41.4%
He3 – 11.2%
n – 33.6%
Figuring energy per particle or number of particles per unit energy
The neutron has an energy of 2.45 MeV
1 MeV of kinetic energy is equal to 1.6x10e6 ergs
A DD fusion neutron has 2.45 x 1.6 x 10e6 = 3.9 x 106 ergs of energy
There are 10e7 ergs in a joule, thus, there are 10e7/3.9 x 10e6 = 2.6 x 10e12 DD neutrons required to represent 1 joule or 1 watt second of energy.
To obtain 1 watt/second of neutrons we must produce 3 watts of DD fusion.
Countrawise,
1 watt of DD fusion only contains 0.336 watts of neutrons or .336 x 2.6x10e12 = 8.7 x10e11 neutrons.
Figuring flux
All calculations are based on an assumed point source of fusion (never the case), in order to simplify calulations.
Only one bit of geometry is needed. The surface area of a sphere is A = 4 pi r^2
Flux is generally considered to be the number of particles, (in this case, neutrons), passing through an area of one square centimeter in one second.
Figure the neutron flux from a 1 watt output DD fusion reaction at a range of 1 meter.
A sphere of 1 meter radius has an area of 4 x pi x 100^2 = 125,664 sq cm
There are 8.7 x10e11 neutron emitted from this reaction
The flux from this reaction at 1 meter would be 8.7 x 10e11 / 125,664 = 7 x10e6 n/sq cm/second.
The difference in dose rate versus total equivalent dose.
Human tissue differs in its absorption of neutrons based on their energy. A plot exists whereby over a span of different neutron energies one might find what flux will produce a given equivalent dose. All dose rates discussed here are once again for the 2.45 MeV DD fusion neutron and will differ for neutrons of other energies.
Dose rate
Radiation dosage is quantified by the ‘rem’ or Roentgen Equivalent Man. Neutrons of varying energy have varying effects and a correction chart for tissue absorption of neutrons of varying energy shows that for DD neutrons about 8.4 neutrons per square cm per second is equivalent to a dose rate of 1millirem/hour.
If a neutron rate meter shows a fusor is putting out a dose rate of 10mrem/hr. Then 84 neutrons are passing through each square cm, each second at the location of the detector probe.
Equivalent Dose
The total dose received over a given period of exposure is the dose rate multiplied by time on exposure. If we are at the probe location in the above example for a period of 8 minutes, we will have received a dose of (8/60) X 10 = 1.3 mrem of fast neutrons.
Another method of arriving at absorbed dose per second can be had when we know the flux at a given point. We suspect that we are receiving a high dose of radiation. We know the flux at our location is 10e5 neutrons per sq cm per sec. How much radiation are we receiving each second? (8.4 x 3600) = 30240 neutrons hitting a square cm to receive 1 mrem/second 10e5 / 3 x 10e4 = 3.3 mrem each second. While not deadly even over minutes of time, a wise man would move out to 10 meters and only receive 3.3/10^2 = 33 microrem/sec. A wiser man, still, would construct a simple paraffinboron shield so that he could stay close to his work.
If we put in 400 watts to our DD fusor and, by calculation, find we are producing 10e5 neutrons per second isotropic emission, what is the energy input to actual fusion energy output ratio? (This a realistic case of a successful but not optimally operated fusor  typical example)
We know, from above, that 1 watt of DD fusion emits 8.7 x 10e11 neutrons isotropically each second. So, we are producing 10e5 / 8.7 x 10e11 = 0.1 microwatt of output energy due to neutrons which we have measured. However, This is only 33% of the total fusion energy. the actual total fusion energy emitted is 3.3 X .1 = .33 microwatt. From this total energy figure we arrive at the requested ratio of 400 / 3.3 x 10 e7 or 1.2 x10e9
Restated we are putting in more than 1.2 billion times as much energy as we are getting out as fusion energy…Or, for each fusion watt output, we must supply to our fusor 1.2 billion watts of electricity.
We find that even the best fusors are still off by a factor of 200 million to one and the average fusor is in the billion to one range. (Nine orders of magnitude improvement to breakeven and 11 orders of magnitude before fusion looks attractive as a power source.
So, we can see that by just using the neutron production as measured from a fusor, we can figure the entire fusion energy produced.
It can't be overemphasized that all the simple calcs above have made specific assumptions designed to make the understanding and order of magnitude calcs easy for the newbie or mathematically handicapped. Any step off the assumptions and the results are skewed and can quickly drift off into fantasy if tight results and data are demanded. Fortunately, in the amateur world, we are not called to task on minutia by overlords looking to hold our feet to the fire. Most are here for the effort and not precision. Still, good scientific investigation, amateur or professional, needs to understand the limits of its data collection accuracy and why those limits exist.
Remember that to work with the simplest of calculations, the number of parameters of an experiment must be limited in number and rigidly fixed and held fixed throughout the experiment. If more parameters are introduced which can change, the complexity of the calculations must increase tremendously to deal these new aspects of the procedure or experiment.
P.S. Thanks to many who helped me make such corrections as needed to all of the above with their comments listed in the posts below. All such corrections are now incorporated above as of 5/22/08.
Richard Hull
All calculations will be carried out to only the nearest tenth part of any value.
DeuteriumDeuterium fusion data
Two reactions occur with about equal probability.
D + D = T (tritium 1.01 MeV) + P (3.02MeV)
D + D = 3He (.82 MeV) + n (2.45 MeV)
NOTE* There is a third reaction possible, but it only occurs once in every ~50,000 DD fusions and that is D+D = 4He + gamma ray.....We will not even consider this reaction in any of our work due to its extreme rarity.
Thus DD fusion is a mixed output of the above two major reactions such that the energy from DD fusion (7.3 MeV) shakes out to the following fractions of the net ongoing reactions.
T – 13.8%
P  41.4%
He3 – 11.2%
n – 33.6%
Figuring energy per particle or number of particles per unit energy
The neutron has an energy of 2.45 MeV
1 MeV of kinetic energy is equal to 1.6x10e6 ergs
A DD fusion neutron has 2.45 x 1.6 x 10e6 = 3.9 x 106 ergs of energy
There are 10e7 ergs in a joule, thus, there are 10e7/3.9 x 10e6 = 2.6 x 10e12 DD neutrons required to represent 1 joule or 1 watt second of energy.
To obtain 1 watt/second of neutrons we must produce 3 watts of DD fusion.
Countrawise,
1 watt of DD fusion only contains 0.336 watts of neutrons or .336 x 2.6x10e12 = 8.7 x10e11 neutrons.
Figuring flux
All calculations are based on an assumed point source of fusion (never the case), in order to simplify calulations.
Only one bit of geometry is needed. The surface area of a sphere is A = 4 pi r^2
Flux is generally considered to be the number of particles, (in this case, neutrons), passing through an area of one square centimeter in one second.
Figure the neutron flux from a 1 watt output DD fusion reaction at a range of 1 meter.
A sphere of 1 meter radius has an area of 4 x pi x 100^2 = 125,664 sq cm
There are 8.7 x10e11 neutron emitted from this reaction
The flux from this reaction at 1 meter would be 8.7 x 10e11 / 125,664 = 7 x10e6 n/sq cm/second.
The difference in dose rate versus total equivalent dose.
Human tissue differs in its absorption of neutrons based on their energy. A plot exists whereby over a span of different neutron energies one might find what flux will produce a given equivalent dose. All dose rates discussed here are once again for the 2.45 MeV DD fusion neutron and will differ for neutrons of other energies.
Dose rate
Radiation dosage is quantified by the ‘rem’ or Roentgen Equivalent Man. Neutrons of varying energy have varying effects and a correction chart for tissue absorption of neutrons of varying energy shows that for DD neutrons about 8.4 neutrons per square cm per second is equivalent to a dose rate of 1millirem/hour.
If a neutron rate meter shows a fusor is putting out a dose rate of 10mrem/hr. Then 84 neutrons are passing through each square cm, each second at the location of the detector probe.
Equivalent Dose
The total dose received over a given period of exposure is the dose rate multiplied by time on exposure. If we are at the probe location in the above example for a period of 8 minutes, we will have received a dose of (8/60) X 10 = 1.3 mrem of fast neutrons.
Another method of arriving at absorbed dose per second can be had when we know the flux at a given point. We suspect that we are receiving a high dose of radiation. We know the flux at our location is 10e5 neutrons per sq cm per sec. How much radiation are we receiving each second? (8.4 x 3600) = 30240 neutrons hitting a square cm to receive 1 mrem/second 10e5 / 3 x 10e4 = 3.3 mrem each second. While not deadly even over minutes of time, a wise man would move out to 10 meters and only receive 3.3/10^2 = 33 microrem/sec. A wiser man, still, would construct a simple paraffinboron shield so that he could stay close to his work.
If we put in 400 watts to our DD fusor and, by calculation, find we are producing 10e5 neutrons per second isotropic emission, what is the energy input to actual fusion energy output ratio? (This a realistic case of a successful but not optimally operated fusor  typical example)
We know, from above, that 1 watt of DD fusion emits 8.7 x 10e11 neutrons isotropically each second. So, we are producing 10e5 / 8.7 x 10e11 = 0.1 microwatt of output energy due to neutrons which we have measured. However, This is only 33% of the total fusion energy. the actual total fusion energy emitted is 3.3 X .1 = .33 microwatt. From this total energy figure we arrive at the requested ratio of 400 / 3.3 x 10 e7 or 1.2 x10e9
Restated we are putting in more than 1.2 billion times as much energy as we are getting out as fusion energy…Or, for each fusion watt output, we must supply to our fusor 1.2 billion watts of electricity.
We find that even the best fusors are still off by a factor of 200 million to one and the average fusor is in the billion to one range. (Nine orders of magnitude improvement to breakeven and 11 orders of magnitude before fusion looks attractive as a power source.
So, we can see that by just using the neutron production as measured from a fusor, we can figure the entire fusion energy produced.
It can't be overemphasized that all the simple calcs above have made specific assumptions designed to make the understanding and order of magnitude calcs easy for the newbie or mathematically handicapped. Any step off the assumptions and the results are skewed and can quickly drift off into fantasy if tight results and data are demanded. Fortunately, in the amateur world, we are not called to task on minutia by overlords looking to hold our feet to the fire. Most are here for the effort and not precision. Still, good scientific investigation, amateur or professional, needs to understand the limits of its data collection accuracy and why those limits exist.
Remember that to work with the simplest of calculations, the number of parameters of an experiment must be limited in number and rigidly fixed and held fixed throughout the experiment. If more parameters are introduced which can change, the complexity of the calculations must increase tremendously to deal these new aspects of the procedure or experiment.
P.S. Thanks to many who helped me make such corrections as needed to all of the above with their comments listed in the posts below. All such corrections are now incorporated above as of 5/22/08.
Richard Hull
Progress may have been a good thing once, but it just went on too long.  Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

 Posts: 590
 Joined: Sun Dec 25, 2005 12:31 am
 Real name:
Re: FAQ  DD fusion  Energy  Flux  Dose
Thank you for the compilation, I think it will be a helpful FAQ. Here are some cases for comparison:
A.
Standing next to a hypothetical 500 W DD fusion neutron source at 1 m distance, you would receive an incapacitating dose of 800 rem within 56 SECONDS. This will not kill you immediately, but surely.
Such exposures can and did happen in criticality accidents, e.g. in Tokaimura 1999. The operators noticed a blue flash and received doses between 8002000 rem in seconds, while the total heat produced was quite small and the fission reaction quenched itself quickly, then restarted again periodically.
B.
If one assumes a DD fusion neutron source with just 1 W fusion power, that translates into 10^12 n/s emitted isotropically. The flux at 1 m will be 8x10^6 n/s/cm^2.
1 mrem corresponds to about 30'000 n/cm^2 for 2.45 MeV DD fast neutrons, over an arbitrary time.
Thus with 1W fusion power you will receive 0.3 rem/s at 1 m distance. This exposure would be survivable, if you noticed quickly enough  within minutes  what is happening.
C.
Considering the homeopathic fusion rates in todays amateur fusors, even the high powered ones, we can expect a neutron production rate of 13x10^6 n/s TIER (total isotropic emission rate) at input power levels of 1KW. A million neutrons per second TIER means 8 n/s/cm^2 at 1 m distance from the fusor's center. This would give the operator or innocent bystanders a dose rate of slightly less than 1 mrem/h. In reality the accumulated dose will be much less, because the fusor is operated for only a few minutes at full power.
In practically all cases, the exposure by xray emissions would be several times higher than the neutron dose. Fortunately, these xrays can be easily shielded.
A.
Standing next to a hypothetical 500 W DD fusion neutron source at 1 m distance, you would receive an incapacitating dose of 800 rem within 56 SECONDS. This will not kill you immediately, but surely.
Such exposures can and did happen in criticality accidents, e.g. in Tokaimura 1999. The operators noticed a blue flash and received doses between 8002000 rem in seconds, while the total heat produced was quite small and the fission reaction quenched itself quickly, then restarted again periodically.
B.
If one assumes a DD fusion neutron source with just 1 W fusion power, that translates into 10^12 n/s emitted isotropically. The flux at 1 m will be 8x10^6 n/s/cm^2.
1 mrem corresponds to about 30'000 n/cm^2 for 2.45 MeV DD fast neutrons, over an arbitrary time.
Thus with 1W fusion power you will receive 0.3 rem/s at 1 m distance. This exposure would be survivable, if you noticed quickly enough  within minutes  what is happening.
C.
Considering the homeopathic fusion rates in todays amateur fusors, even the high powered ones, we can expect a neutron production rate of 13x10^6 n/s TIER (total isotropic emission rate) at input power levels of 1KW. A million neutrons per second TIER means 8 n/s/cm^2 at 1 m distance from the fusor's center. This would give the operator or innocent bystanders a dose rate of slightly less than 1 mrem/h. In reality the accumulated dose will be much less, because the fusor is operated for only a few minutes at full power.
In practically all cases, the exposure by xray emissions would be several times higher than the neutron dose. Fortunately, these xrays can be easily shielded.
 Carl Willis
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 Real name: Carl Willis
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Re: FAQ  DD fusion  Energy  Flux  Dose
Hi Richard,
Thanks for the straightforward FAQ on this topic.
If I were to add anything, it might be to simply emphasize to readers (these assumptions are already explained, but I know some people will neglect them!) that the 8.4 n / cm^2 / sec / mrem / hr is specific to ~2.5 MeV monoenergetic neutrons, and the flux and doseatadistance calculations assume an isotropic distribution of products particles. Deviations from the assumptions mean those simplifications no longer apply.
Also a minor semantics correction: what you've labeled the "absorbed dose" should be called the "equivalent dose" (absorbed dose is measured in rads or Gray).
Carl
Thanks for the straightforward FAQ on this topic.
If I were to add anything, it might be to simply emphasize to readers (these assumptions are already explained, but I know some people will neglect them!) that the 8.4 n / cm^2 / sec / mrem / hr is specific to ~2.5 MeV monoenergetic neutrons, and the flux and doseatadistance calculations assume an isotropic distribution of products particles. Deviations from the assumptions mean those simplifications no longer apply.
Also a minor semantics correction: what you've labeled the "absorbed dose" should be called the "equivalent dose" (absorbed dose is measured in rads or Gray).
Carl
 Richard Hull
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 Joined: Fri Jun 15, 2001 1:44 pm
 Real name: Richard Hull
Re: FAQ  DD fusion  Energy  Flux  Dose
Thanks to Carl and Wilifried. All corrections are now included. The proper terminology is important for us all.
Richard Hull
Richard Hull
Progress may have been a good thing once, but it just went on too long.  Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.
 Steven Sesselmann
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 Joined: Thu Aug 11, 2005 1:50 am
 Real name: Steven Sesselmann
 Location: Sydney  Australia
 Contact:
Re: FAQ  DD fusion  Energy  Flux  Dose
Richard,
Thanks for another informative FAQ, which will be very useful for the many newbies that have joined lately.
Although I agree with your calculations, I hope you don't mind if I pick on the semantics a bit.
Reading your post it comes across to me, that Q (input vs. output energy) depends on the neutron energy only. If this were the case then 66% of the D+D reaction energy would disappear into nowhere.
One must of course consider the full 7.3 Mev per two fusions, when calculating efficiency.
It makes the figures look just a little less depressing..
Steven
Thanks for another informative FAQ, which will be very useful for the many newbies that have joined lately.
Although I agree with your calculations, I hope you don't mind if I pick on the semantics a bit.
Reading your post it comes across to me, that Q (input vs. output energy) depends on the neutron energy only. If this were the case then 66% of the D+D reaction energy would disappear into nowhere.
One must of course consider the full 7.3 Mev per two fusions, when calculating efficiency.
It makes the figures look just a little less depressing..
Steven
http://www.gammaspectacular.com  Gamma Spectrometry Systems
https://www.researchgate.net/profile/Steven_Sesselmann  Various papers and patents on RG
https://www.researchgate.net/profile/Steven_Sesselmann  Various papers and patents on RG
 Richard Hull
 Moderator
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 Joined: Fri Jun 15, 2001 1:44 pm
 Real name: Richard Hull
Re: FAQ  DD fusion  Energy  Flux  Dose
Thanks for the correction. I refigured to the total fusion energy, which was the actual question posed. This was an important correction. Thanks.
I personally fail to see how one is at all bouyed by the 3X correction. as the 9 order magnitude remains to breakeven and 11 order to electrical grid attraction. 3 times nothing is just a bit more of nothing.
I have acknowledged the corrections and added a date of the corrected FAQ
Richard Hull
I personally fail to see how one is at all bouyed by the 3X correction. as the 9 order magnitude remains to breakeven and 11 order to electrical grid attraction. 3 times nothing is just a bit more of nothing.
I have acknowledged the corrections and added a date of the corrected FAQ
Richard Hull
Progress may have been a good thing once, but it just went on too long.  Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

 Posts: 590
 Joined: Sun Dec 25, 2005 12:31 am
 Real name:
Re: FAQ  DD fusion  Energy  Flux  Dose
A few more words on the dose definition might be in order. This can be a tricky matter, also due to the number of historical radiation units, which adds to the confusion.
"Absorbed dose" here is the amount of energy deposited in matter by ionizing radiation. (1Gy=1J/kg)
"Equivalent dose" is the absorbed dose multiplied by a quality factor to reflect the biological effect, depending on the type of radiation. Fast neutrons are the most damaging of all.
dose rate = dose per time unit (e.g. Gy/h)
dose = dose rate x time (e.g. Gy)
equivalent dose = dose x quality factor Q, depending upon radiation. (e.g Sv, mrem)
So we can compare absorbed dose rate and equivalent dose rate, but not vs. equivalent dose.
That's like comparing speed to distance.
For fast neutrons of 2.45 MeV, Q is somewhere between 10 and 20.
The actual biological effectiveness is smaller and difficult to measure. Q is defined for radiation protection purposes, so it is exaggerated on the safe side.
"Absorbed dose" here is the amount of energy deposited in matter by ionizing radiation. (1Gy=1J/kg)
"Equivalent dose" is the absorbed dose multiplied by a quality factor to reflect the biological effect, depending on the type of radiation. Fast neutrons are the most damaging of all.
dose rate = dose per time unit (e.g. Gy/h)
dose = dose rate x time (e.g. Gy)
equivalent dose = dose x quality factor Q, depending upon radiation. (e.g Sv, mrem)
So we can compare absorbed dose rate and equivalent dose rate, but not vs. equivalent dose.
That's like comparing speed to distance.
For fast neutrons of 2.45 MeV, Q is somewhere between 10 and 20.
The actual biological effectiveness is smaller and difficult to measure. Q is defined for radiation protection purposes, so it is exaggerated on the safe side.
 Richard Hull
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 Joined: Fri Jun 15, 2001 1:44 pm
 Real name: Richard Hull
Re: FAQ  DD fusion  Energy  Flux  Dose
I, not being a professional member of the physics community nor writing professional papers for publication within it, do hereby refuse to use SI units. I was born and raised a CGS guy and so I will perish as one.
I hated MKS, but have found a particularly acute loathing deep within myself for SI. There is but one SI unit that I will speak in, if forced, and that is the very reasonable, Becquerel. How such a tiny, logical and reasonable unit made the SI cut I can't imagine. Some one asleep at the switch, I guess.
My beef is one of scale and seemingly constant and needless revisionism. We work in a laboratory of sorts, both pros and amateurs. Such facilities are people sized and project components within them mostly are of the order of the human hand as a median size. Really tiny experiments can go to atoms and large stuff we can handle or guide and control go to many meters in size.
CGS covers the median, laboratory sized stuff.
MKS was designed for some sort of great out of doors laboratory. (falling safes and pianos, lobing artillary shells and designing rockets.
SI seems to look at a unit magnetic field greater than most any permenant magnet air gap can achieve and other larger than lab or even life units. Admittedly, a number of the old MKS stuff moved over, some with the names changed to confuse the innocent.
I realize that, slowly with each old fossils death, SI will become the norm...................Or will it?.......Will Phd. Carl Willis be forced, at 57, to adapt to the OUS (Otovian Units System)?!!
Don't chuckle too hard. In college I had to deal with one physics class that still talked english in poundals and slugs and then hit a Chemistry and Electrical math class that spoke only in CGS. While picking up additional classes later at the Univesity of Florida in the late 60's, it was all MKS. Now it is SI. Welcome to a world where your units are like the shifting sands and light bulbs forever seem to go off in the heads of the annointed that wind up keeping the confusion gas jet turned up full blast.
I understand your points but will not include the SI units in my posting. Folks that chose to handle the finer points of radiation physics in SI need to get a modern textbook and study it. This effort was a quick rinse for newbs looking for easy speak.
I mentioned in the post above that the neutron flux selected came from a neutron energy flux to tissue dose conversion plot to create a rem value for a specific flux at a specific energy. I have already spoken to the biological effectiveness rating multipliers in a previous radiation FAQ.
Richard "CGS" Hull
I hated MKS, but have found a particularly acute loathing deep within myself for SI. There is but one SI unit that I will speak in, if forced, and that is the very reasonable, Becquerel. How such a tiny, logical and reasonable unit made the SI cut I can't imagine. Some one asleep at the switch, I guess.
My beef is one of scale and seemingly constant and needless revisionism. We work in a laboratory of sorts, both pros and amateurs. Such facilities are people sized and project components within them mostly are of the order of the human hand as a median size. Really tiny experiments can go to atoms and large stuff we can handle or guide and control go to many meters in size.
CGS covers the median, laboratory sized stuff.
MKS was designed for some sort of great out of doors laboratory. (falling safes and pianos, lobing artillary shells and designing rockets.
SI seems to look at a unit magnetic field greater than most any permenant magnet air gap can achieve and other larger than lab or even life units. Admittedly, a number of the old MKS stuff moved over, some with the names changed to confuse the innocent.
I realize that, slowly with each old fossils death, SI will become the norm...................Or will it?.......Will Phd. Carl Willis be forced, at 57, to adapt to the OUS (Otovian Units System)?!!
Don't chuckle too hard. In college I had to deal with one physics class that still talked english in poundals and slugs and then hit a Chemistry and Electrical math class that spoke only in CGS. While picking up additional classes later at the Univesity of Florida in the late 60's, it was all MKS. Now it is SI. Welcome to a world where your units are like the shifting sands and light bulbs forever seem to go off in the heads of the annointed that wind up keeping the confusion gas jet turned up full blast.
I understand your points but will not include the SI units in my posting. Folks that chose to handle the finer points of radiation physics in SI need to get a modern textbook and study it. This effort was a quick rinse for newbs looking for easy speak.
I mentioned in the post above that the neutron flux selected came from a neutron energy flux to tissue dose conversion plot to create a rem value for a specific flux at a specific energy. I have already spoken to the biological effectiveness rating multipliers in a previous radiation FAQ.
Richard "CGS" Hull
Progress may have been a good thing once, but it just went on too long.  Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.
 Chris Bradley
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Re: FAQ  DD fusion  Energy  Flux  Dose
I do tend to agree  I like to use centimetres, but then usually confuse myself throughly by combining it with kg!!!
Depends on the exercise, I guess. I always used to build virtual computer RF models using units of imperial feet. The reason; the speed of light = 1 foot/ns (near enough). Very convenient!
Depends on the exercise, I guess. I always used to build virtual computer RF models using units of imperial feet. The reason; the speed of light = 1 foot/ns (near enough). Very convenient!
 Chris Bradley
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Re: FAQ  DD fusion  Energy  Flux  Dose
Good point on Xrays. Could/should this FAQ also cover Xray emissions? When most likely, dose, most appropriate protection for a given dose (and/or best value?  lead's getting pricey these days!), etc..
best regards,
Chris MB.
best regards,
Chris MB.