Need some help with maths
- Thread starter Davros
- Start date
Here's the wikipedia entry on E-notation
https://en.wikipedia.org/wiki/Scientific_notation#E_notation
https://en.wikipedia.org/wiki/Scientific_notation#E_notation
Silent_Buddha
Legend
8.0776087e+33 means the decimal point has been moved 33 places to the left and it is rounded to the nearest 8th significant figure. Expanded this would be
80776087000000000000000000000000000 [edit]fixed![/edit]
Obviously you can see the rounding error involved.
But it's good enough for the shorthand notation so someone doesn't have to type (or write) out the entire thing. Similar to how Pi being typically represented as 3.14 or 3.14159 is rounded and obviously not 100% accurate but it's good enough for a lot of purposes.Scientific notation is also basically required on most calculators as they just can't display the entire number.
Regards,
SB
Last edited:
Silent_Buddha
Legend
Whoops, yup, I sure did. See how typing out a longer number makes it more error prone? (At least that's my excuse)
Regards,
SB
See how typing out a longer number makes it more error prone? (At least that's my excuse)Regards,
SB
Relativity. Closing speed is actually (u + v) / (1 + (uv / c^2))
Where u and v are the speeds of the two objects.
When u and v are small (compared to c), uv/c^2 is very small, so the closing speed is almost equal to the sum of the speeds. This means that just adding the speeds together is close enough for any sensible car/train/etc.
yes I know, I was doing it as a mental exercise, Im trying to teach myself Special Relativity
The real formula is in the pic
where v1 is the speed of car 1, v2 is the speed of car 2 and C is the speed of light
as cjo points out the speed is so close to 200kmh theres no point in using the correct formula with everyday speeds unless you need to know the speed accurate to a trillionth of a kph
Go to the 1:00 mark
ps: I find it funny that a video that says the way youve been doing it is just an approximation then approximates the speed of light
makes me wonder if they used the correct value of C in calculating the speed of the sheep
If your trying to be 100% correct then no.
The real formula is in the pic
where v1 is the speed of car 1, v2 is the speed of car 2 and C is the speed of light
as cjo points out the speed is so close to 200kmh theres no point in using the correct formula with everyday speeds unless you need to know the speed accurate to a trillionth of a kph
Go to the 1:00 mark
ps: I find it funny that a video that says the way youve been doing it is just an approximation then approximates the speed of light
makes me wonder if they used the correct value of C in calculating the speed of the sheep
Last edited:
Which part? If it's nutball's suggestion, it works as follows:
If both entities are moving at half of the speed of light, you would expect them to be moving towards each other at the speed of light.
This means that two objects, each travelling at half of the speed of light, end up moving towards each other at 80% of the speed of light.
Billy Idol
Legend
Say v1 = x*c and v2=y*c
Then you get as the result
(x+y)/(1+xy)*c
Typically, in your everyday life, x<<1 and y<< 1 and with this, the above result is approx
~(x+y)*c as one would expect for instance in the case of cars, or even super sonic aircrafts,
as (1+xy) is ~ 1, as x*y is much smaller than 1.
The most interesting case is actually x=y=1, traveling with speed of light.
In this case, the result is:
(1+1)/(1+1*1)*c = 1*c
showing that the speed of light is an upper limit: no information can travel faster than c.
(Suppose you are passenger in a hypothetical train going with speed of light. Now you stand up from your seat and start walking in the direction of travel...what is your actual speed?)
Btw, this is not math, but physics
Then you get as the result
(x+y)/(1+xy)*c
Typically, in your everyday life, x<<1 and y<< 1 and with this, the above result is approx
~(x+y)*c as one would expect for instance in the case of cars, or even super sonic aircrafts,
as (1+xy) is ~ 1, as x*y is much smaller than 1.
The most interesting case is actually x=y=1, traveling with speed of light.
In this case, the result is:
(1+1)/(1+1*1)*c = 1*c
showing that the speed of light is an upper limit: no information can travel faster than c.
(Suppose you are passenger in a hypothetical train going with speed of light. Now you stand up from your seat and start walking in the direction of travel...what is your actual speed?)
Btw, this is not math, but physics
Your speed versus the moving train? Because I've always taken the liberty to reinterpret 'general relativity' as that there is no observable movement unless you have two objects one of which moves, and there is no time unless there is one movement to measure and another movement to measure with.
So if you are saying that two objects move towards each other at the speed of 100km/h, then given no further context the closing speed should be exactly 200km/h. Speed of light is only of influence on observing this movement and if gravitation of earth and other factors have an impact on the closing speed then the two cars weren't actually moving towards each other at the specified speed.
I realize that I should probably watch that video
Edit: and after watching that video, I have not changed my mind. if two cars go at each other at 200km/h, but they are traveling on a curve (earths surface) then they don't go straight towards each other anymore and so the closing speed changes ever so slightly. And the sheep on the truck that drives on a curve walks on a slightly higher curve, so walking the same speed over a slightly longer distance would calculate to an ever so slightly lower walking speed?
I should probably go to sleep now
So if you are saying that two objects move towards each other at the speed of 100km/h, then given no further context the closing speed should be exactly 200km/h. Speed of light is only of influence on observing this movement and if gravitation of earth and other factors have an impact on the closing speed then the two cars weren't actually moving towards each other at the specified speed.
I realize that I should probably watch that video
Edit: and after watching that video, I have not changed my mind.
if two cars go at each other at 200km/h, but they are traveling on a curve (earths surface) then they don't go straight towards each other anymore and so the closing speed changes ever so slightly. And the sheep on the truck that drives on a curve walks on a slightly higher curve, so walking the same speed over a slightly longer distance would calculate to an ever so slightly lower walking speed? I should probably go to sleep now
Last edited:
c is a constant, the speed of light. The example was if both objects were travelling at half of the speed of light. If you look at my image, when you feed two objects travelling half of the speed of light in to the equation, the result is a closing speed of 80% of the speed of light.
Last edited:
Thats not an equation for the relative speed. Thats where he's telling you what speeds v1 and v2 ARE.
V1 is the same speed as V2.
V1 is half the speed of light.
V2 is half the speed of light.
Being a pedant here but the speed of light squared will in m^2/s^2 not m/s. >8->
Absolutely correct from the perspective of either object! To either one of them, the other object is in every way approaching at 80% of the speed of light.
But let's think of the implied external observer who measures one object moving north at 50% of the speed of light and the other object on a collision course moving south at 50% the speed of light. What's the closure speed? In this case, the simple linear speed addition is CORRECT and you can say that the objects are approaching each other at the speed of light. It would even be correct if each one was moving at 90% of the speed of light, and you would properly say they were approaching each other at 180% of the speed of light. You're using a speed to measure a seperation distance which is decreasing over time at a rate of 1.8 c. There's no paradox: no object or information is moving faster than light.
Similar threads
