# Newtons to Watts to Joules

1. Mar 5, 2015

### shott92

so basically i have calulated some forces in newtons at one second intervals
i have
time - 1, 2, 3, 4, 5, ----- 5 seconds total
distance - 3, 12, 27, 48, 72 ------ 162 meters total

force = 1804, 1822, 1849, 1866, and 101 newtons
energy= 5411, 16'369, 27'736, 39'187, 2432 joules ---- calculated using Force*distance -
power = 5411, 16'369, 27'736, 39'187, 2432 watts ---- calculated by energy/time (1 second intervals)

but this all goes to junk when i go to work out the total cumulative figures for the run, because
i either just add them up and joules is thus the same as watts =7442 newtons, and 91135 joules and watts
or i add up the total force and the multiply by total distance divide the total time, 7442 netwons , *162 meters = 1205604 joules, and /5 second = 241120.8 watts

then i thought i would sus it by using the si base unit principle, force =m*d/t^2, energy = m*d^2/t^2, power = m*d^2/t^3 but that didnt acount for the other forces that was not dependant on mass so it ended up far out anyways... can someone help please thanks in advance cheers

Last edited: Mar 5, 2015
2. Mar 5, 2015

### mathman

The difference is mathematical. Adding them gives the correct result.

3. Mar 5, 2015

### Staff: Mentor

I am not certain what you mean here. The only way that you ever add forces is if there are multiple forces acting on the same object at the same time. You never add forces on different objects and you never add forces at different times.

4. Mar 5, 2015

### shott92

i mean adding the accumulated forces, im attempting to create a drive cycle so i have to consider, not only the Mass*Acceleration but tractive force, rolling resistance, gradient, areodynamic forces and drivetrain losses etc. as well these forces i have added up taking essentially points every second (if this makes sense)
so at the first second all these forces added together equalled 1804 newtons at the second (2nd) second 1822 newtons,
i have done this for 300 seconds, i just took a sample of 5 for the sake of this post,
i then was trying to calculate the total amount of energy and power used, to do this i added the force from all the seconds together in my example above it would have been,
7442 newtons, then i worked out the power from this...
is that any clearer at all ??

5. Mar 5, 2015

### Staff: Mentor

You should never do that.

6. Mar 5, 2015

### shott92

ah right so what is the appropriate method to calcualte the total amount of joules and the total power requirede for whole run??
is my first method that the total amount of watts and joules both being the same correct? becuase this doesnt seem right

7. Mar 5, 2015

### jbriggs444

The total energy for the run is the sum of the energies for each incremental bit of the run.

The total power for the run is a nonsensical number. The average power, on the other hand, is a useful figure and can be computed as total power divided by total time.

8. Mar 5, 2015

### shott92

right i think i get you the total energies are for each incremental bit of the run, and as such the total energy would be
5411+16'369+27'736+39'187+2432= 91135 newton*meters
but the power would be an average so as such 91135/5 = 18227 joules per second (watts)
and if anybody knows can someone show me somewhere i can verify it because im just getting confused by one saying yup add them, another no never add them and you saying average it out (your answer seems to fit with the numbers i want the best but i dont really have a clue :'( )

9. Mar 5, 2015

### shott92

also if thats the case how do i work out kilowatt hours from there or can that be converted from joules? they are both units of energy right?

10. Mar 5, 2015

### jbriggs444

DaleSpam is correct (he usually is). Adding up the forces would be wrong. Dale did not say anything one way or the other about power.

Kilowatt hours is 1000 watts times one hour = 1000 joules per second times 3600 seconds = 3.6 million Joules. And yes, it is a unit of energy.

11. Mar 5, 2015

### Staff: Mentor

You are asking about several different things. If you calculate the energy of each segment then you can add those up to get a total energy. The total energy divided by the total time is the average power.

You cannot add up the forces over time to get a total force. So any calculation where you do that is wrong, and that is why "it all goes to junk" as you said originally.

12. Mar 6, 2015

### shott92

awesome i think i understand now so the forces cant be added so there is no accumulative force (is there a derivative or something that is used instead?? because it seemed really logical to me that there be some form of total or average or some figure that should be here)
the power can be added to give joules as 91135 newton*meters
and the total energy over total time give average power, 18227 joules per second (watts)
so from this i guess that i can get to kw*h from 18.277 * 3600 = 65797.2 kwh
or from the joules where it would be 91.135 * (3600/5) = 65617.2 kwh
this all makes much more sense and its no longer "going to junk" so thank you very much guys your help has been extremely usefull and is greatly appriciated

13. Mar 6, 2015

### jbriggs444

You cannot add power to get energy. The units are not correct.

What you could do is to multiply the power for the first second by the duration of the first interval (1 second) to get the energy for the first interval. You could do the same to get the energy for the second. And for the third, fourth and fifth. You could add those energies together to get the total energy.

But you already knew the energy for the first, second, third, fourth and fifth intervals. Computing it based on power would be silly.

Yes.

If you are producing energy at a rate of 18 kilo Joules per second on average and you do this continually for 3600 seconds, how many kilo Joules do you get?

14. Mar 6, 2015

### shott92

sorry i did mean energy here :/

ah right 18.227 kilo joules * 3600 = 65797.2 kilo joules or 65.8 mega joules right ?? so is it 18kwh?? :/

15. Mar 6, 2015

### jbriggs444

Yes

16. Mar 6, 2015

### Staff: Mentor

In this type of problem you really need to pay close attention to the units and understand the mathematical operations that are physically meaningful.

So let's consider force. It has units of N. Now let's say that you apply a constant force of 6 N for 10 s, and you measured the force every second. Now, you might think that you could simply add those samples to get 60 N, but what if you sampled every 2 s? Then you would only get 30 N. That doesn't make any sense, you didn't change the force at all, you still have a constant 6 N for 10 s, but you changed your "sum" just by changing your sampling frequency. That is not useful.

So instead, what you can calculate is something called "impulse" which is the force times the time it was applied. Hopefully you can see that this is a quantity whose sum does not change based on the sampling frequency. But notice that the units are different, instead of N the units are s*N.

17. Mar 6, 2015

### sophiecentaur

Adding up forces over a period of time doesn't produce a useful quantity. That would imply that a brick on top of a wall was, somehow, having a different effect, merely by sitting there with its 10N weight force acting downwards for five minutes or five million years. The work done, as the wall compresses microscopically, would be force times the displacement but has nothing to do with how long it took to happen.
You have to make up valid equations in Science, random inclusion of variables doesn't produce meaningful answers.

18. Mar 6, 2015

### A.T.

What is your actual goal? What are you trying to find out?

19. Mar 6, 2015

Staff Emeritus
This.

It's like saying it's 80 degrees in Miami, 30 degrees in Boston and 50 degrees in Seattle, in total 160 degrees.

20. Mar 7, 2015

### sophiecentaur

The mean temperature may be of some interest - and so might the mean force but that is a separate issue from the 'total' temperature or force.

21. Mar 7, 2015

### shott92

bascially the whole point of my exersize it to creat a drive cycle that can be used and adapted for different condition, vehicle weights and powers, to find out the neccersary motor size then the battery size for the vehicle to be able to do the drive cycle around 18 times (its a 5 min drive cycle so 1.5 hours in total)

the motor size i have been trying to calculate by using all the force that act upon the vehicle at any given second - then working out the power from there, and also they energy and power used for the drive cycle in fairness i have been able to achive the desired result without the need for accumulative force but i just kinda felt that it was the natural way to go and cant quite get my head round why it doesnt work as in my mind its not like the temp in boston, miami and seattle its like the rain in boston on day one the rain in boston on day two and the rain in boston on day three, and should therefor be addable as well as find out mean presipiation amounts ect.

22. Mar 7, 2015

### Staff: Mentor

Did you read post 16? I explained why there. Maybe my explanation was not clear.

23. Mar 7, 2015

### sophiecentaur

Perhaps it could have been put more simply? The (Implied) maths may have put him off.
I wonder, what did the OP hope to achieve by adding the force every second for 30 seconds (as an example)? If you added the forces every 5 seconds, 10 seconds or 1/1000 seconds you would get different answers - telling you what?
Basically, if you invent a measurement then it has to relate to something.
Work, Impulse, Momentum and Kinetic Energy are all quantities that are of some use in these circumstances and allow you to progress with design and analysis. What would this "total force" figure do for you? What would it tell you?

This is not the first time, iirc, that the "total force`' idea has come up.

24. Mar 7, 2015

### shott92

i infact had not and only saw the later 3 posts, i do appologise, it actually became really obvious once i saw the principle i was trying to achive, im really sorry i missed it and carried on with my question, i am please to say i am now having a good look into impluse, something i haven't actually covered before and it is kind of overwelming to behonest but my simple understanding of it ATM, is that
where momentum is = m*v
and force is = m*a or m*v/t
impulse is essentially = m*j or m*v/t/t ?
anyhow i will look into this more, and dale thanks for your help and once again sorry for missing your excellent post :/