You totally rock, mate. Thanks for your great and clear posts and please keep sharing. You have a very happy Aussie the other side of the planet waiting for more posts!
Good formula.Useful when converting from a longer distance to a shorter chest radiography when patient is unable to stand and unable to sit and still be able to maintain the same film density
Selisa Freeman Here's the link on amazon... you can follow me on twitter and facebook also. I usually do raffles and coupons, but it's less than $5 for Kindle: www.amazon.com/Becoming-Radiologic-Technologist-Jeremy-Enfinger-ebook/dp/B008HVCVS8/ref=sr_1_1?ie=UTF8&qid=1390966928&sr=8-1&keywords=becoming+a+radiologic+technologist
I've seen that question going around a lot... sounds like a great one. Sorry about the late reply to your comment here. For some reason, I'm not getting notifications anymore, but I'm checking my settings.
Intensity is not inversely proportional. It's inversely related to the square of the distance. Being inversely proportional would mean if you double your distance, you halve the intensity, which would be incorrect. This formula shows the relationship (and how to calculate for changes) between radiation intensity and the distance from its source. Hope that helps... I've been thinking of a more visual way to explain this for a future video, but this is how you'd do the math.
rlalko l I'm sorry you didn't find it helpful... I showed the steps I did in order to make it easy to understand... the work can be cut into fewer steps with a calculator, but I really only used 4 steps - after you plug in valued from your formula, 1. reduce 2. square the distance values 3. cross-multiply 4. divide so that "x" is by itself on one side of the equal sign. Keep in mind the reason I show all of these steps is because I look for each of these when I grade student exams to ensure they know how to do the work. After a little practice, you should be able to do these relatively quickly in your head as a practicing technologist.
You totally rock, mate. Thanks for your great and clear posts and please keep sharing. You have a very happy Aussie the other side of the planet waiting for more posts!
Good formula.Useful when converting from a longer distance to a shorter chest radiography when patient is unable to stand and unable to sit and still be able to maintain the same film density
Thanks... there's more to come. I'm going to be posting at least once a week, but I'll try to get in some more. Stay tuned....
Thank you so much sir!! You really do helped me understanding this example.
Thank you so much! You may have turned the semester around for me.
Haha glad it helped
sir, how do you know how much you will reduced, and why did you reduced it?
Headache saver!!! Thanks!
Thanks so much this helped so much. I understand now what to look for in the problem to find out what formula to use... keep them coming!
No problem... I'll see what I can work up in the future :-)
Awesome. I'm a student in the x - ray program and was wondering how can I get your book. ..I have a Kindle.
Selisa Freeman Here's the link on amazon... you can follow me on twitter and facebook also. I usually do raffles and coupons, but it's less than $5 for Kindle: www.amazon.com/Becoming-Radiologic-Technologist-Jeremy-Enfinger-ebook/dp/B008HVCVS8/ref=sr_1_1?ie=UTF8&qid=1390966928&sr=8-1&keywords=becoming+a+radiologic+technologist
Thanks. I'm ordering as we speak.
Selisa Freeman Great! Thanks for the support. Please let me know what you think once you've read it.
Thank you sir! Great vid!
can you please do videos covering all of radiology math formulas. I've been take St Catherine's test and their math problem are like 5 steps
Do you have examples of problems you're struggling with? I haven't done a video for a long time, but shouldn't be too tricky.
I would like to know if you have any lessons on how the change in mAs kVp etc will affect density contrast and recorded detail
There are several posts on my website dealing with these... you can use google or the search bar on the site at topicsinradiography.com
I wanted to know about 15 percent rule with an example and possible questions
I've seen that question going around a lot... sounds like a great one. Sorry about the late reply to your comment here. For some reason, I'm not getting notifications anymore, but I'm checking my settings.
please expose time calculation explain.
Super helpful, thank you!
Sure!
intensity is inversely proportional to the distance, i dont actually get what you mean by what you just did
Intensity is not inversely proportional. It's inversely related to the square of the distance. Being inversely proportional would mean if you double your distance, you halve the intensity, which would be incorrect. This formula shows the relationship (and how to calculate for changes) between radiation intensity and the distance from its source. Hope that helps... I've been thinking of a more visual way to explain this for a future video, but this is how you'd do the math.
Thank you SO much!!!
Sure!
Ty, ty so much
No problem :-)
I got 15mas .. man i have my exam Thursday and I am struggling with the math sadly. Can anyone help me?
Thanks a lot very helpful
Thank you!
No problem!
@WhySkinCare : No problem :-)
it should be mAs 2 over mAs 1 = SID2 squared over SID1 squared
Scared for rad program cuz of math lol.
why did you make it so hard, you added to many steps... less steps are easier not everyone learns the same. i did this in 6 steps
rlalko l I'm sorry you didn't find it helpful... I showed the steps I did in order to make it easy to understand... the work can be cut into fewer steps with a calculator, but I really only used 4 steps - after you plug in valued from your formula, 1. reduce 2. square the distance values 3. cross-multiply 4. divide so that "x" is by itself on one side of the equal sign. Keep in mind the reason I show all of these steps is because I look for each of these when I grade student exams to ensure they know how to do the work. After a little practice, you should be able to do these relatively quickly in your head as a practicing technologist.