# Struggling to convert Newtons to watts

1. Jun 9, 2015

### ncrenwick28

Hello.

I am not an engineer but would really appreciate some help with a calculation for my PhD in biomedical sciences. I have a new exercise ergometer (see picture) and have been able to obtain the force applied (newtons) to the load cells located underneath each foot pedal at 1hz. A motor moves the the 2 foot plates back and forth at 23rpm whilst the subject pushes against them. The velocity and distance remain constant so only the force applied (Newtons) varies.
I have completed one study with participants exercising on this new ergometer however for publishing purposes i need to convert my current integrated newtons profile (n.min) into Watts.
I have measured the distance from the pivot point to the load cell as 54cm and the distance the foot plate moves back and forth as 16cm.
This is probably a really easy calculation but my medical head can't quite grasp it.

Any help would be greatly appreciated

Nick
http://www.btetech.com/images/eccentron-beauty-pre.jpg [Broken]

Last edited by a moderator: May 7, 2017
2. Jun 9, 2015

### Svein

Power (W) = Force (N) times Distance (m) divided by time (s). So if your subject has produced a force of 100N over a distance of 20m in 30 seconds, the power is (100⋅20)/30 W = 67W.

3. Jun 9, 2015

That's an average power, of course.

4. Jun 9, 2015

### ncrenwick28

Thankyou Svein.
I have seen this calculation but am a bit confused. The machines motor is actually moving the pedals not the subjects so therefore should distance be an integral part of the equation? If so, would i use the distance from the load cell the the pivot point ( a constant 0.54m) or the distance the pedals move in the z axis (0.16m)?

Cheers
Nick

5. Jun 9, 2015

### gmax137

Just checking: you do know that this calculation (force x distance pedals moved) will provide the work per time (power) being applied to the pedals, ie, not the work per time being done by the test subject, right?

6. Jun 9, 2015

### ncrenwick28

Oh ok that makes sense. Do you know how i would be able to calculate the power (W) being done by the test subject then?
Nick

7. Jun 9, 2015

### ncrenwick28

Anyone have any idea?
I'm guessing the extra power required by the machines motor to overcome the participants force applied would equal the participants power output, however the power output from the machine is unknown.

8. Jun 9, 2015

### jack action

First, you cannot convert Newtons into Watts. Those are units that represents different things (force and power). It is literally comparing apples and oranges.

The notion of (mechanical) power implies a force $F$ in motion, and motion implies a velocity $v$. The true definition of power is $F \cdot v$. Those are two vectors and it is a dot product. In rotation the definition is based on torque $\tau$ and angular velocity $\omega$, power is $\tau \cdot \omega$.

So the power received by the load cell is the force applied to the load cell times the speed of the cell in the direction of the load. The force can vary and the velocity can vary. Power is instantaneous and can be measured at one point in space. If velocity is constant and you can calculate the average force through a path, or if the force is constant and you can calculate the average velocity through a path, then you can calculate the average power. But if both force and velocity vary, multiplying both averages will not necessarily give you the average power.

In your case the load cell velocity varies tremendously as the pedal goes back and forth (going from 0 to vmax, back to 0 then -vmax and 0 again). But if your rotational motion is at constant angular velocity (23 rpm), you can calculate the average torque produced by the load cell to that rotational motion and then average power can be more easily calculated with $\tau \cdot \omega$.

Note that if the cell is not moving, no work is done, hence the power is zero. For example, you could have put the load cell on the seat and still measure the same load (the user pushes the pedals but also the seat at the same time -> equal and opposite reaction) but since the seat doesn't move, there is no power to measure on that end.

If your pedals are moved by a motor with no load on the pedals, the motor will have a certain power output (to fight back friction and such). If the user pushes the pedals, then the pedals and motor should accelerate. If they don't, it means that the motor has somehow increased its torque to produce the amount of power that the user has generated (in addition to the friction power).

9. Jun 10, 2015

### ncrenwick28

Hi Jack.
I think i may have worked it out, or at least an average power as the velocity of the pedals vary.
I measured the degree displacement of the pedals and converted to radians, then worked out how many radians per second (angular velocity)based on 23rpm (one rpm being the movement backwards and forwards of the pedal)
Then i just multiply the current torque value with my calculated angular velocity, Sound right??

2 pedal movements for 1rpm so.. 0.227*2 = 0.454
1rpm = 60/23 = 2.61s
1 radian per second = 0.454/2.61 = 0.174

If the load cell output is 150N the power (W) = 150*0.174 = 26.1W

Thanks again

10. Jun 10, 2015

### jack action

This is wrong, as the units don't even match.

First, rpm means revolution per minute. What you are describing is cycle per minute, which is not the same. Here how it goes:

Now, what you need to know is the actual distance traveled within one cycle (let's call it d), which would be θR, where R is the radius of the arc described by one pedal (in meter) and θ is the total angular distance traveled (previously calculated).

If you have 23 cycles/min then the average speed in m/min is 23*d. Divide it by 60 to convert it into m/s.

Then you can multiply the load cell (in Newtons) by that average speed (in m/s) to obtain the average power in Watts. Remember that - because we actually use the average speed - we assume the force on the load cell is constant throughout the motion. If it is not, it may not correctly represents the average power.

11. Jun 10, 2015

### gmax137

I'm cautioning against trying to relate measured work on the device (from force x distance on the pedals) to energy expenditures in biologic or metabolic terms (kcal/hr or similar). The difficulties have been discussed here many times, for example in the "weightlifting" threads.

12. Jun 11, 2015

### ncrenwick28

Ok thanks everyone. my radius of the arc is 0.54 metres. so if the applied torque was 150N that gives me 14.1W. Correct?
Cheers
Nick