- #1
phoenix_1326
- 3
- 0
To start with, I have never been in a physics class, nor am I an engineer. I am a computer programmer, and I love the science channel, so at best I am a bad amateur in this field of study. That being said...
I was thinking about calculations that could be used in games like Angry Birds, and I thought back to Einstein's E = MC2. I understand this formula is used to calculate the TOTAL energy released when mass is annihilated, but I thought there might be a way of using this formula to calculate the energy used when say a baseball is in flight. I had what I thought was a eureka moment when I came up with E = MC2 / A where A is the actual MPH the baseball is thrown at.
I looked up the mass of a baseball: 142 g. As I understand it, that makes M = .142
C = ~186,282
The pitcher can throw at 95 mph, so A = 95
As such...
E =.142(186,2822) / 95 = .142(34,700,983,524) / 95 = 4,927,539,660.408 / 95 = 51,868,838.53061053
E = ~5.27 Joules...or something like that
This of course is where I have run into a major issue. I have no idea if I am even looking at this problem the right way, or how it translates into even the digital world (let alone the real world where...according to other posts on this site...the "springy-ness" of the material comes into play as well as the size of the material vs. the size of the point of impact...ok, the head ache just hit).
Can someone break this down into a "general and laymen" understanding of what I'm looking at? I would greatly appreciate it...and I know my wife will as well...she knows I'm going to obsess on this for the next month or until I figure it out enough to satisfy me.
I was thinking about calculations that could be used in games like Angry Birds, and I thought back to Einstein's E = MC2. I understand this formula is used to calculate the TOTAL energy released when mass is annihilated, but I thought there might be a way of using this formula to calculate the energy used when say a baseball is in flight. I had what I thought was a eureka moment when I came up with E = MC2 / A where A is the actual MPH the baseball is thrown at.
I looked up the mass of a baseball: 142 g. As I understand it, that makes M = .142
C = ~186,282
The pitcher can throw at 95 mph, so A = 95
As such...
E =.142(186,2822) / 95 = .142(34,700,983,524) / 95 = 4,927,539,660.408 / 95 = 51,868,838.53061053
E = ~5.27 Joules...or something like that
This of course is where I have run into a major issue. I have no idea if I am even looking at this problem the right way, or how it translates into even the digital world (let alone the real world where...according to other posts on this site...the "springy-ness" of the material comes into play as well as the size of the material vs. the size of the point of impact...ok, the head ache just hit).
Can someone break this down into a "general and laymen" understanding of what I'm looking at? I would greatly appreciate it...and I know my wife will as well...she knows I'm going to obsess on this for the next month or until I figure it out enough to satisfy me.