One of the more difficult topics for first year students is the correlation between kV and density. Once you think you have this concept down, including the 15% rule and the subsequent lab experiments, this topic gets revisited a number of times throughout the entire x-ray program, and some very good questions about kV, mAs, and density typically arise. For instance:
Does the number of photons increase as kV increases?
Well, yes... let me explain first by clarifying that the number of electrons produced at the cathode does not increase - that is controlled only by mA while the duration of production is controlled by the time.
Now, let's say I have a technique of 65 kV and 10 mAs for a knee x-ray. A certain number of electrons are converted to x-ray photons at the anode during that exposure, and then a certain number of primary photons are converted to secondary and tertiary photons and so forth when they interact with the patient until one of two things will happen to all photons:
1 - they will leave the patient as scatter or expose the film
2 - they will lose potential difference and become absorbed in the patient
Now let's focus on the photons in the latter category... when we increase kV, we know that more photons reach the film because they have increased energy to penetrate the patient, but something else happens. There will still be a percentage of photons that will be absorbed in the patient, but it will not be as high of a percentage as the 60 kV exposure. You will now, at 75 kV for instance, have more energy even in the photons that are absorbed, to ionize tissue before those photons deposit all energy into tissue.
To clarify, let's say we have a characteristic interaction between a primary x-ray photon and an atom of carbon in the patient. We'll also say that this photon carried 75 kV of potential difference. The binding energy of the k shell for carbon is .28 keV, so we're left with a secondary x-ray photon of 74.72 keV with the ability to produce the same reaction approximately 267 more times (75/.28) before it is absorbed and each reaction produces more photons that will be absorbed. If the same series of reactions occurred with the 65 kV exposure, then you would only have a possible 232 of these identical interactions before the photon's energy is absorbed.
Keep in mind that this is only one example of a photon's interaction with matter, and there is always that randomness applied to how they react. If you haven't studied compton, photoelectric, or characteristic interactions yet, don't feel bad if you didn't understand the last paragraph. Just remember that when a photon interacts with matter, other photons are typically produced, or electrons are ejected that can ionize adjacent atoms as well, and increasing kV will increase the number of interactions that occur before the photons are absorbed.
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Does the number of photons increase as kV increases?
Well, yes... let me explain first by clarifying that the number of electrons produced at the cathode does not increase - that is controlled only by mA while the duration of production is controlled by the time.
Now, let's say I have a technique of 65 kV and 10 mAs for a knee x-ray. A certain number of electrons are converted to x-ray photons at the anode during that exposure, and then a certain number of primary photons are converted to secondary and tertiary photons and so forth when they interact with the patient until one of two things will happen to all photons:
1 - they will leave the patient as scatter or expose the film
2 - they will lose potential difference and become absorbed in the patient
Now let's focus on the photons in the latter category... when we increase kV, we know that more photons reach the film because they have increased energy to penetrate the patient, but something else happens. There will still be a percentage of photons that will be absorbed in the patient, but it will not be as high of a percentage as the 60 kV exposure. You will now, at 75 kV for instance, have more energy even in the photons that are absorbed, to ionize tissue before those photons deposit all energy into tissue.
To clarify, let's say we have a characteristic interaction between a primary x-ray photon and an atom of carbon in the patient. We'll also say that this photon carried 75 kV of potential difference. The binding energy of the k shell for carbon is .28 keV, so we're left with a secondary x-ray photon of 74.72 keV with the ability to produce the same reaction approximately 267 more times (75/.28) before it is absorbed and each reaction produces more photons that will be absorbed. If the same series of reactions occurred with the 65 kV exposure, then you would only have a possible 232 of these identical interactions before the photon's energy is absorbed.
Keep in mind that this is only one example of a photon's interaction with matter, and there is always that randomness applied to how they react. If you haven't studied compton, photoelectric, or characteristic interactions yet, don't feel bad if you didn't understand the last paragraph. Just remember that when a photon interacts with matter, other photons are typically produced, or electrons are ejected that can ionize adjacent atoms as well, and increasing kV will increase the number of interactions that occur before the photons are absorbed.
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This makes good sense :) More photons are produced, but not in the primary beam. The additional photons that are produced from photoelectric and compton effect within the body are lower energy because the binding energy in the K shell of atoms such as carbon, hydrogen, oxygen, and nitrogen is much lower (lower atomic number elements). This is why tungsten is used in the anode target. It has a high enough atomic number and density to produce diagnostic quality radiation, as well as a melting point high enough to withstand the heat of the reaction. (I hope this is correct... I'm still learning)
ReplyDeleteThis makes good sense :) More photons are produced, but not in the primary beam. The additional photons that are produced from photoelectric and compton effect within the body are lower energy because the binding energy in the K shell of atoms such as carbon, hydrogen, oxygen, and nitrogen is much lower (lower atomic number elements). This is why tungsten is used in the anode target. It has a high enough atomic number and density to produce diagnostic quality radiation, as well as a melting point high enough to withstand the heat of the reaction. (I hope this is correct... I'm still learning)
ReplyDeleteSounds good... I'm not sure what semester you're in, but you may not need to know this yet. It will come back around though.
ReplyDeleteSounds good... I'm not sure what semester you're in, but you may not need to know this yet. It will come back around though.
ReplyDeleteHi,
ReplyDeleteI'm a second semester Rad student. We are currently on topics of kvp, density and contrast, etc. I have the concept of what each does but I'm having a real problem with the word problems we are doing. x-ray taken at .07 sec @ 200 mA and 80 kVp the contrast was low. Calculate a technique to give a high contrast at the 200 mA station. I would really appreciate it if you could explain this to me in simple terms. I'm just not grasping it.
Thanks
Hi,
ReplyDeleteI'm a second semester Rad student. We are currently on topics of kvp, density and contrast, etc. I have the concept of what each does but I'm having a real problem with the word problems we are doing. x-ray taken at .07 sec @ 200 mA and 80 kVp the contrast was low. Calculate a technique to give a high contrast at the 200 mA station. I would really appreciate it if you could explain this to me in simple terms. I'm just not grasping it.
Thanks
A couple of fundamentals must have already been studied for this problem: You should know that doubling mA (and leaving time alone) will double your density/mAs. Conversely, if you double your time (and leave mA alone), it will have the same effect. Second, you have to apply the 15% rule and maintain density. In order to increase contrast, you need to decrease kVp by 15%. 15% of 80 = 12, and 80-12=68.
ReplyDeleteNow, with 68 kVp, you will have higher contrast, but you've just cut your radiographic density in half. You now have to double your mAs to maintain your original density. Since you have to keep the 200 mA station constant, your only option is to double your time from .07 sec to .14 sec.
So your new technique will be 68 kVp, 200 mA and .14 seconds. You will have higher contrast while maintaining density and your 200 mA selection.
I hope this helps!
A couple of fundamentals must have already been studied for this problem: You should know that doubling mA (and leaving time alone) will double your density/mAs. Conversely, if you double your time (and leave mA alone), it will have the same effect. Second, you have to apply the 15% rule and maintain density. In order to increase contrast, you need to decrease kVp by 15%. 15% of 80 = 12, and 80-12=68.
ReplyDeleteNow, with 68 kVp, you will have higher contrast, but you've just cut your radiographic density in half. You now have to double your mAs to maintain your original density. Since you have to keep the 200 mA station constant, your only option is to double your time from .07 sec to .14 sec.
So your new technique will be 68 kVp, 200 mA and .14 seconds. You will have higher contrast while maintaining density and your 200 mA selection.
I hope this helps!
Thanks so much. It did help. It's going to be a long semester, I may need your help again. Thanks again.
ReplyDeleteThanks so much. It did help. It's going to be a long semester, I may need your help again. Thanks again.
ReplyDeletehow do you know when to use the inverse square law as opposed to the direct square law?
ReplyDeleteThe inverse square law is used to calculate intensity (usually in mR), which should decrease the farther away the patient is from the tube.
ReplyDeleteWhen you are increasing or decreasing SID and you need to find out what new mAs to select, then you should use the direct square law.
Basically, if you double your distance/SID, then your intensity is 1/4 the original value (inverse square), so that means you will need to use 4x the mAs to "maintain density" (direct square/density maintenance).
The inverse square law is used to calculate intensity (usually in mR), which should decrease the farther away the patient is from the tube.
ReplyDeleteWhen you are increasing or decreasing SID and you need to find out what new mAs to select, then you should use the direct square law.
Basically, if you double your distance/SID, then your intensity is 1/4 the original value (inverse square), so that means you will need to use 4x the mAs to "maintain density" (direct square/density maintenance).
I am a first quarter Rad student and we are learning about the inverse square law. I just can't understand it. Could you please give me an example and explain it to me? I have a test next week and I am freaking out!!!!!
ReplyDeleteThanks!!!!!!!
I am a first quarter Rad student and we are learning about the inverse square law. I just can't understand it. Could you please give me an example and explain it to me? I have a test next week and I am freaking out!!!!!
ReplyDeleteThanks!!!!!!!
Here's an example problem:
ReplyDeleteA radiographic technique produces an exposure of 200 mR at an SID of 100 cm. What would the exposure be at an SID of 150 cm?
200/x = (150)2/(100)2
Application of the inverse square law concludes the answer to be 88.8 mR
Here's an example problem:
ReplyDeleteA radiographic technique produces an exposure of 200 mR at an SID of 100 cm. What would the exposure be at an SID of 150 cm?
200/x = (150)2/(100)2
Application of the inverse square law concludes the answer to be 88.8 mR
DO NOT EVER GO AWAY...PLEASE
ReplyDeleteI am a first year rad tech student who was drowning even with two texts to explain this stuff, until I found your blog...GOD BLESS YOU for being here. I am just trying to digest what I am reading now, but will most certainly have questions/comments in the future. THANK YOU FOR THIS SITE!!!!
DO NOT EVER GO AWAY...PLEASE
ReplyDeleteI am a first year rad tech student who was drowning even with two texts to explain this stuff, until I found your blog...GOD BLESS YOU for being here. I am just trying to digest what I am reading now, but will most certainly have questions/comments in the future. THANK YOU FOR THIS SITE!!!!
O.k, if I am understanding this correctly...
ReplyDelete1) If I was a practicing Rad Tech and wanted to know what my technique would change to if I was changing my SID, then I would use the Direct Square Law: example, a radiographic technique is set at 100 mAs at 180cm and I want to know what the new mAs would be at 90cm,
(90cm/180cm)squared x 100 mAs = new mAs RIGHT?
If not please, explain.
2) Under what circumstances would a Rad Tech use the Inverse Square Law? Why?
O.k, if I am understanding this correctly...
ReplyDelete1) If I was a practicing Rad Tech and wanted to know what my technique would change to if I was changing my SID, then I would use the Direct Square Law: example, a radiographic technique is set at 100 mAs at 180cm and I want to know what the new mAs would be at 90cm,
(90cm/180cm)squared x 100 mAs = new mAs RIGHT?
If not please, explain.
2) Under what circumstances would a Rad Tech use the Inverse Square Law? Why?
Thanks for your comments Al... sorry it's taking so long to reply. It's been a very busy week!
ReplyDeleteYou are correct in your utilization of the formula for #1... you know that you are cutting your distance in half, so you will need 1/4 the mAs.
For #2, there will be questions about radiation intensity that you will need to use this for. It will always ask (instead of mAs) about R or mR and the relationship between intensity and distance... basically, the farther away you are from the radiation source, the less the intensity is.
We use this mostly in quality assurance when we do testing of the equipment for calibration, routine maintenance servicing, QC, etc. If you are a current student, these QA/QC topics are typically covered later in the program.
I hope this helps!
Thanks for your comments Al... sorry it's taking so long to reply. It's been a very busy week!
ReplyDeleteYou are correct in your utilization of the formula for #1... you know that you are cutting your distance in half, so you will need 1/4 the mAs.
For #2, there will be questions about radiation intensity that you will need to use this for. It will always ask (instead of mAs) about R or mR and the relationship between intensity and distance... basically, the farther away you are from the radiation source, the less the intensity is.
We use this mostly in quality assurance when we do testing of the equipment for calibration, routine maintenance servicing, QC, etc. If you are a current student, these QA/QC topics are typically covered later in the program.
I hope this helps!
Hi!
ReplyDeleteI am a student radiographer and I have a question :
What is the main factor under the radiographer control effecting image contarst, image density and image sharpenes??
Is it by increasing the kV?
thanks in advance
Hi!
ReplyDeleteI am a student radiographer and I have a question :
What is the main factor under the radiographer control effecting image contarst, image density and image sharpenes??
Is it by increasing the kV?
thanks in advance
Traditionally, the main controller of contrast is kVp... as you increase kVp, contrast decreases (inverse relationship).
ReplyDeleteThe main controller of density is mAs. As you double mAs, your optical density should double (direct relationship).
Sharpness is a tricky term. I believe you may be referring to resolution, or the ability to distinguish adjacent structures (my apologies if this is not what you are referring to). The main thing you can control resolution with is your focal spot size selection. The smaller the focal spot, the higher the resolution should be on your final radiograph.
You should be learning these concepts within the first two semesters of your Radiography Program. They can become quite confusing because although they can primarily affect contrast/density/resolution, they can affect other attributes of your image as well.
Traditionally, the main controller of contrast is kVp... as you increase kVp, contrast decreases (inverse relationship).
ReplyDeleteThe main controller of density is mAs. As you double mAs, your optical density should double (direct relationship).
Sharpness is a tricky term. I believe you may be referring to resolution, or the ability to distinguish adjacent structures (my apologies if this is not what you are referring to). The main thing you can control resolution with is your focal spot size selection. The smaller the focal spot, the higher the resolution should be on your final radiograph.
You should be learning these concepts within the first two semesters of your Radiography Program. They can become quite confusing because although they can primarily affect contrast/density/resolution, they can affect other attributes of your image as well.
Hi- I am taking a Intro to Radiology before I can apply for the program and I am so confused on how to answer these math problems. the intensity is 90mR at 40 in. What is it at 72in? Our teacher will not give us any examples and does not explain anything, I really hope you can help.
ReplyDeleteThanks
Hi- I am taking a Intro to Radiology before I can apply for the program and I am so confused on how to answer these math problems. the intensity is 90mR at 40 in. What is it at 72in? Our teacher will not give us any examples and does not explain anything, I really hope you can help.
ReplyDeleteThanks
how does doubling the mAs effect density? we are doing an experiment where we take 3 exposure of a wrist. one with 60 kV and 2.5 mAs, second one with 60 kv and 5 mAs, the last one is at 60 kV 10 mAs.
ReplyDeleteHi,
ReplyDeleteIm a 1st year radiography student.
I would like to know how do we calculate mAs for 1uGy to image receptor?
Or if there's any relevant texts i could refer to?
please help :)
Hi,
ReplyDeleteIm a 1st year radiography student.
I would like to know how do we calculate mAs for 1uGy to image receptor?
Or if there's any relevant texts i could refer to?
please help :)
what happens to the x-ray tube when mAs is set??
ReplyDeleteand ehat happens to the x-ray tube when kVp is set?
what happens to the x-ray tube when mAs is set??
ReplyDeleteand ehat happens to the x-ray tube when kVp is set?
After the desired mA and time are set, nothing really "happens" inside the x-ray tube until the exposure switch is pressed... most switches are two-stage.
ReplyDeleteWhen the switch is pressed down half-way, two things happen. The rotor begins accelerating at the anode side of the tube. You'll notice a delay between the time you begin pressing the switch and when the machine will allow you to take the exposure; that's because it is waiting for complete and constant acceleration of the rotor, which spins the tungsten anode disc for heat dissipation. The other thing that occurs is thermionic emission - boiling off of electrons at the selected filament on the cathode side. A pre-designated number of electrons will be boiled off, controlled by the mA station you select.
When the exposure switch is fully depressed (after the ready light comes on), two things now happen here. Your kilovoltage is applied to the circuit, allowing the electron cloud from the cathode to be pushed across the gap to the anode disc, changing the energy to x-rays. Also, a timer circuit prevents this from occurring longer (or shorter) than desired based on your overall mAs setting. It terminates the application of kVp to the circuit when the desired mAs value has been reached, giving you the desired density at that mAs value. When AEC is used, the timer circuit is terminated when an appropriate level of exposure reaches the selected ion chambers.
After the desired mA and time are set, nothing really "happens" inside the x-ray tube until the exposure switch is pressed... most switches are two-stage.
ReplyDeleteWhen the switch is pressed down half-way, two things happen. The rotor begins accelerating at the anode side of the tube. You'll notice a delay between the time you begin pressing the switch and when the machine will allow you to take the exposure; that's because it is waiting for complete and constant acceleration of the rotor, which spins the tungsten anode disc for heat dissipation. The other thing that occurs is thermionic emission - boiling off of electrons at the selected filament on the cathode side. A pre-designated number of electrons will be boiled off, controlled by the mA station you select.
When the exposure switch is fully depressed (after the ready light comes on), two things now happen here. Your kilovoltage is applied to the circuit, allowing the electron cloud from the cathode to be pushed across the gap to the anode disc, changing the energy to x-rays. Also, a timer circuit prevents this from occurring longer (or shorter) than desired based on your overall mAs setting. It terminates the application of kVp to the circuit when the desired mAs value has been reached, giving you the desired density at that mAs value. When AEC is used, the timer circuit is terminated when an appropriate level of exposure reaches the selected ion chambers.
By increasing the generator kV for a given exposure, you DO NOT alter the amount of electrons released at the kathode - that is determined by the mA setting. But you DO increase the amount of X-ray photons released in the anode. Because the released electrons have more kinetic energy (due to higher kV), they will interact more often with anode atoms (Compton effect) and therefore will release more X-ray photons, with higher average kinetic energy.
ReplyDeleteNote that the average photon energy in the x-ray beam is about 1/3 the generator output.
E.G. a 90kV generator setting will result in an average photon energy of about 30keV.
By increasing the generator kV for a given exposure, you DO NOT alter the amount of electrons released at the kathode - that is determined by the mA setting. But you DO increase the amount of X-ray photons released in the anode. Because the released electrons have more kinetic energy (due to higher kV), they will interact more often with anode atoms (Compton effect) and therefore will release more X-ray photons, with higher average kinetic energy.
ReplyDeleteNote that the average photon energy in the x-ray beam is about 1/3 the generator output.
E.G. a 90kV generator setting will result in an average photon energy of about 30keV.
Hello
ReplyDeleteSomebody let me know how I can memorize or learn Kv & mAs or mA & sec for different parts of body. e.g. knee, pelvis, skull, limbs , ...
Please! because whenever I go to hospital, after positioning of patient I must tell somebody to check exposure parameters and this one depress me! It's a pleasure to me that you have such a great website.
Hello
ReplyDeleteSomebody let me know how I can memorize or learn Kv & mAs or mA & sec for different parts of body. e.g. knee, pelvis, skull, limbs , ...
Please! because whenever I go to hospital, after positioning of patient I must tell somebody to check exposure parameters and this one depress me! It's a pleasure to me that you have such a great website.
You should really be operating kVp and mAs values within an "optimal range of exposures" which can vary from patient to patient. Most x-ray programs will offer methods for technique chart production in the latter half of your courses. Until that point, just try to observe body habitus, pathology, presence/absence of fluid, and exposure indicators in relationship to the techniques you are using. You will begin to notice patterns in these methods, and eventually, you will be very comfortable with them and can even contribute to formation of technique charts once you get to that portion of your x-ray program.
ReplyDeleteI sincerely hope you are still here! I am a new rad student and just as I think I am understanding the 15% rule, the teacher poses a question that is confusing to me. As I understand it, increasing the kVp by 15% will roughly double the image density; decreasing the kVp by 15% will reduce the image density to roughly one-half. If I want to produce an image with more contrast and half the density, would I simply decrease the kVp by 15%? Would the original mAs technique remain the same?
ReplyDeleteYes... still here. You are correct - if you only lower the kVp 15%, you will reduce density by 1/2 and the contrast will increase (any time you decrease kVp).
ReplyDeleteRemember though - you need adequate density to evaluate contrast. If the overall image is too light, then your not be able to properly evaluate contrast. Your statement above assumes that adequate density will be produced with your change in kVp.
Yes... still here. You are correct - if you only lower the kVp 15%, you will reduce density by 1/2 and the contrast will increase (any time you decrease kVp).
ReplyDeleteRemember though - you need adequate density to evaluate contrast. If the overall image is too light, then your not be able to properly evaluate contrast. Your statement above assumes that adequate density will be produced with your change in kVp.
Hi ..
ReplyDeleteThanks alot for all effort that u have done !
I am a 1st. master student , i have a certin project which is about (The effect of KvP on Dinsty &caontrast). The problem is I dont know how to start !!
Can u help me in the beginning??
Thank you sooo much :)
Hi ..
ReplyDeleteThanks alot for all effort that u have done !
I am a 1st. master student , i have a certin project which is about (The effect of KvP on Dinsty &caontrast). The problem is I dont know how to start !!
Can u help me in the beginning??
Thank you sooo much :)
You should probably start with explaining the difference between density and contrast. Please feel free to email me any questions you have... topicsinradiography@gmail.com thanks for reading!
ReplyDeleteI'm a second year student who is struggling with technique changes still!
ReplyDeleteI'm preparing an image analysis for an end of semester paper and I can't find the answer to this...
an image of a hand was done with a fixed tech 50@2.5 mAs, done in error using a grid and then repeated. What can I say about the use of grid on small part? Image was washed out...of no diagnostic value..and what tech could be used if tech used grid properly?
I'm a second year student who is struggling with technique changes still!
ReplyDeleteI'm preparing an image analysis for an end of semester paper and I can't find the answer to this...
an image of a hand was done with a fixed tech 50@2.5 mAs, done in error using a grid and then repeated. What can I say about the use of grid on small part? Image was washed out...of no diagnostic value..and what tech could be used if tech used grid properly?
Well, you shouldn't use a grid on parts under 10cm, so your density (or exposure to the plate) would be insufficient. I don't believe there would be a proper use of a grid on a hand.
ReplyDeletehye
ReplyDeleteim third year rad student..now im doing research on the relationship between mA and second on image quality.the objectives is to determine the density of the image by changing the mA and second.but i dont know how to start.please help.
thanks.
hye
ReplyDeleteim third year rad student..now im doing research on the relationship between mA and second on image quality.the objectives is to determine the density of the image by changing the mA and second.but i dont know how to start.please help.
thanks.
hey, only discvrd ths site ystrdy bt i hv learnt so much. im in my 3rd yr n im writing a research paper n i need help with a theme.
ReplyDeleteThanks for reading and commenting... I want to just say publicly that while I don't mind answering specific questions about particular topics, I need to draw the line somewhere - and that line resides where I'm doing the project for you:
ReplyDeleteRegarding the question about how to start research on how changes in mA and time affect density: this is a first year rad student concept, and it sounds like an excellent activity for improving critical thinking skills using a topic you should already know the basic relationships between by now. I feel like providing my input would hinder the point of the assignment. You could research "popular ways to start a research paper" on google and see what comes up.
The most recent question about finding a theme for a research paper: That's really up to you. What I would suggest is that you find a topic that interests you, then it will be easy to write about and fun to learn more about. Brainstorm some topics that you think you might like, and then do some quick internet research to make sure there is plenty of available research out there for the purpose of your project. One you have defined the scope of your research, and are having trouble understanding the reason for a particular topic, please, by all means, feel free to ask me a specific question. I appreciate that you have come to my blog for assistance, but I believe it would be unethical for me to provide you that level of assistance. If I were your instructor I would want you to be the initiator of your subject material.
As an instructor myself, I ensure that my own students are given the duty, as well as the academic freedom, to find their own topics. I usually assign research papers so that the student is required to learn much more detail about a topic presented in class to improve their understanding of the topic, as well as to give them practice in the methodology behind proper research.
hye....i am radiofrapher stdnt...can i ask smthg about the rltshp between distance sid and exposure factor (kvp and mas) in cxr? why we used different exposure factors between PA chest on the grid bucky (180 cm) and direct exposre (100 cm) such as portable or ap supine? is it because of inverse law? does the grid play a role in deciding exposure factr...how about the pt dose. which one is more higher.tq
ReplyDelete@an nyek: I think I understand your question... two factors play a role in technique changes doing portables. If you use a grid, you will need to compensate for exposure to your image receptor if the ratio varies from your bucky. Also, the inverse square law explains why we need to adjust our mAs at different distances (use density maintenance formula). The photons are not as concentrated at larger SID's, so we need to increase the quantity of photons to ensure that the appropriate amount will reach the image receptor. When comparing dose, it would help to have more information like SID, kVp, mAs, grid ratio (or non-grid), etc.
ReplyDeleteHelp I'm having grid confusion! satisfactory radiograph made w/out grid at 72in sid and 8 mas. if distance changed to 40in and 12.1 grid ratio what is the new mas value to maintain original density?
ReplyDeleteAlright... two-step problem. First, use density maintenance formula to get your new mAs at 40", which is 2.469, but we'll call it 2.5. Then plug in 2.5 as your "mAs 1" value into the grid conversion formula... the conversion factor for a 12:1 grid is 5, so 2.5 mAs x 5 = 12.5 mAs for your new exposure to maintain exposure/density.
ReplyDeleteWow, you are truly gifted at explaining these topics. I wish our instructors had the same command that you demonstrate with these concepts...
ReplyDeleteDo you know about turns ratios? I've got a question that's been bugging me, maybe it would help to bring it to you. Thanks,
Shant
Wow, you are truly gifted at explaining these topics. I wish our instructors had the same command that you demonstrate with these concepts...
ReplyDeleteDo you know about turns ratios? I've got a question that's been bugging me, maybe it would help to bring it to you. Thanks,
Shant
I am in 3rd semester(summer), will be done by the 31st of july. I am having trouble with relating focal spot size/blur, subject contrast( like when we speack of image contrast,we know higher contrast image will be of a hand and lover contrast will be a chest- inversly ralated but how is a subject contrast directly related and how do i not get mixed up with those)
ReplyDeletewe had problems in one of our tests, I couldn't solve it, would you kindly look at it?
-a radiograph is taken using 75KVP@20 Mas,which change in tech would increase contrast but maintain the same density?
I chose = 65kvp@20 Mas(because density doesnt change but since we need higher contast we would have to lower out kvp by 15%.
The answer was 65kvp@40Mas accorsing to the teacher,she failed to explain well.
- a radiograph taken using 65 kvp@10 Mas is too light.which tech would double the OD while producing a wider scale of contrast?
my answer-55kvp@20 Mas(bc the radiograph is too light, we need more blacks/white so decreasing kvp by 15% will increase my density,wouldn't it?
Answer was- 75 KvP@ 10mas...how???
with AEC,the exposure is terminated when the optimum ______ is reached.
Mas or OD?? i got confused bc they both are the same, chose Mas, answer was OD...??
I have been struggling with these questions and i can't figure out why...please explain them to me...would really appreciate it.
thanks!
I am in 3rd semester(summer), will be done by the 31st of july. I am having trouble with relating focal spot size/blur, subject contrast( like when we speack of image contrast,we know higher contrast image will be of a hand and lover contrast will be a chest- inversly ralated but how is a subject contrast directly related and how do i not get mixed up with those)
ReplyDeletewe had problems in one of our tests, I couldn't solve it, would you kindly look at it?
-a radiograph is taken using 75KVP@20 Mas,which change in tech would increase contrast but maintain the same density?
I chose = 65kvp@20 Mas(because density doesnt change but since we need higher contast we would have to lower out kvp by 15%.
The answer was 65kvp@40Mas accorsing to the teacher,she failed to explain well.
- a radiograph taken using 65 kvp@10 Mas is too light.which tech would double the OD while producing a wider scale of contrast?
my answer-55kvp@20 Mas(bc the radiograph is too light, we need more blacks/white so decreasing kvp by 15% will increase my density,wouldn't it?
Answer was- 75 KvP@ 10mas...how???
with AEC,the exposure is terminated when the optimum ______ is reached.
Mas or OD?? i got confused bc they both are the same, chose Mas, answer was OD...??
I have been struggling with these questions and i can't figure out why...please explain them to me...would really appreciate it.
thanks!