Disclaimer: unlike rowing boats, ergs differ fundamentally from one manufacturer to another. Any useful discussion of the physics of ergometers will inevitably have a bias towards the model most commonly used (i.e. Concept), which also happens to be the one that I am most familiar with.
| (2.1) | ω = v / r |
Where mass m occurs in equations for translational systems, in rotational systems this is replaced by the moment of inertia, I:
| (2.2) | I = ∑ r 2 δm |
which is the summation of each component δm of the overall mass at each radius r.
Linear momentum (m x v) is replaced by angular momentum J:
| (2.3) | J = I ω |
In a translational system, Newton's 2nd Law equates force F to a rate of change of momentum d(m v) / dt, or a constant mass m multiplied by a linear acceleration a = dv / dt. In a rotational system, force is replaced by torque, T, and Newton's 2nd Law becomes:
| (2.4) | T = d(I ω) / dt = I ( dω / dt ) = I b |
where a constant moment of inertia I has been assumed (i.e. a rigid body) and b is the angular acceleration.
Work W and Power P (=rate of doing work) are the same in both translational and rotational systems, but derived from different expressions:
| Linear | Rotational | |
|---|---|---|
| (2.5) | W = F x | W = T θ |
| (2.6) | P = F v | P = T ω |
where x is the linear distance moved, and θ (=Greek letter theta) is the rotation angle. Eq. (2.5) assumes that force and torque remain constant during the work - strictly these should be integrations: ∫dW = ∫Fdx = ∫Tdθ
| (3.1) | m / t = a ω |
where a is some constant dependent on the vent-settings (e.g. open vents = increased air flow for given rpm = larger value of a). This same air leaves the system with an outward velocity also proportional to the flywheel speed, so acquiring a kinetic energy proportional to the square of the flywheel speed. A mass m of air passing through the system therefore gains energy:
| (3.2) | E = ½ d m ω2 |
where d is some constant for a given flywheel design. Putting together Eqs. (3.1) and (3.2), this implies that the flywheel is losing energy to the air at a rate (=power dissipation) proportional to the cube of the angular velocity:
| (3.3) | P = E / t = ( E / m ) ( m / t ) = k ω3 |
where k (=½d a) comprises the constants from the previous two equations. This power dissipation appears as a 'drag torque' D, proportional to the square of the angular velocity ω (from Eq. 2.6, but see Note 1):
| (3.4) | D = k ω2 |
| (4.1) | P = F v |
The handle linear velocity is related to the flywheel angular velocity (Eq. 2.1) by the radius of the cog (or pulley) r:
| (4.2) | v = ω r |
The cog-size also determines the relationship between the applied force F and the resulting torque T on the flywheel:
| (4.3) | T = F r |
If the flywheel rotates at a constant angular velocity, the applied torque T (Eq. 4.3) must balance the average drag torque (Eq. 3.4):
| (4.4) | F r = k ω2 |
So the power P required from the rower (Eqs. (4.1) and (4.2)) is:
| (4.5) | P = F ω r = k ω3 |
Suppose the cog radius is changed from r to r1. To maintain the same flywheel speed the applied force F will also have to be changed to F1 (using Eq. 4.4):
| (5.1) | F1r1 = k ω2 = F r |
Rearranging this gives the new force F1:
| (5.2) | F1 = F r / r1 |
But, for a given flywheel speed ω, changing the cog also changes the handle speed from v to v1 (using Eq. 4.2):
| (5.3) | v1 / r1 = ω= v / r |
which can be rearranged to give the new handle speed v1:
| (5.4) | v1 = v r1 / r |
So using F1 from Eq. (5.2) and v1 from Eq. (5.4), the power P1 required to maintain the original speed with the new cog is:
| (5.5) | P1 = F1 v1 = ( F r / r1 ) ( v r1 / r ) = F v = P |
i.e. same as the old power. Putting this into words: although you can maintain a certain flywheel speed with less force using a larger cog, you will need to apply this force quicker so that the power (=force x velocity) required actually remains the same.
It is exactly the same as riding a bicycle in a different gear, or changing the gearing, or length, of an oar.
Changing the air-flow into the erg means that the constant a in Eq. (3.1) is altered. Since the constant k in Eq. (3.4) is proportional to a, this changes the drag torque:
| (6.1) | D = k1 ω2 |
If the same flywheel speed ω is to be maintained, the power required P1 (from Eq. 2.6) is then given by
| (6.2) | P1 = k1 ω3 |
Substituting for ω3 from Eq. (3.3),
| (6.3) | P1 = ( k1 / k) P |
so, unlike the cog-change, this time the power is different.
There are also other, unintentional, ways in which the damping can change: changes in friction in the bearings with age, proximity to a wall or other ergometers, air-pressure, viscosity... For these reasons the ergometer computers do not measure the vent/lever position to determine the damping torque via the constant a, but instead use a more direct method, explained in the next section.
| (7.1) | D = -I ( dω / dt ) |
If the rate of change of angular velocity dω/dt (i.e. deceleration) is measured, and the moment of inertia I is known (presumed the constant, and the same for all flywheels of the same model), the damping torque D can be calculated.
However, D itself is a function of angular velocity ω (Eq.3.4) so a more fundamental parameter is the 'drag factor' k
| (7.2) | k = - I ( dω / dt ) (1 / ω2 ) = I d(1/ω) / dt |
where k can either be calculated rotation by rotation using the first expression and averaged, or just once for the whole recovery phase using the second expression. The calculated 'Drag Factor' k can be displayed on the Concept monitor (units 10-6 N m s2) and something similar called 'Resistance Factor' is displayed on the Rowperfect monitor (I'm afraid I don't know the units).
By measuring the damping, the ergometer will automatically compensate for any of the following:Things that are not compensated are:
- Opening/Closing the vents to increase/reduce resistance
- Changes in friction on the flywheel bearings with time
- Changes in air pressure, density, viscosity etc.
- Environmental factors such as proximity to walls or other ergs
- Changes in the chain friction
- Changes in the tension of the return mechanism
- Manufacturing variations flywheel moments of inertia (probably negligible)
- Changes in flywheel moments of inertia (unlikely with the solid flywheels)
The first two effects are uncompensated because they have no influence on the damping, the second two are uncompensated because they affect the moment of inertia which is assumed to be just a fixed number in the damping calculation. As mentioned in section 4, the apparent change in resistance from changing the cog is purely a 'gearing' effect (i.e. unrelated to damping) so no compensation is required to account for it.
| (8.1) | T - D = I ( dω / dt ) |
The energy dE supplied by the rower to turn the flywheel through an angle dθ is therefore given by (Eq. 2.5):
| (8.2) | dE = T dθ | = I ( dω / dt ) dθ + D dθ |
| = I ( dω / dt ) dθ + k ω2 dθ |
where all the terms are either constant (I), directly measured (ω,t,θ), or assumed constant from derivation during the previous recovery phase (k, see Eq.7.2). The Rowperfect has an additional term representing work done against the tension of the shock cord.
The average supplied power per stroke is then simply obtained by dividing the energy per stroke E by the time taken for each stroke cycle t
| (8.3) | P = E / t |
See Note (2).
| (9.1) | P = k ω3 = c u3 |
So it is natural to relate fan rotational velocity ω to indicated linear velocity u, and the number of rotations θ to the indicated distance covered s:
| (9.2) | u = ( k / c )1/3 ω |
| (9.3) | s = ( k / c )1/3 θ |
For fixed-damping ergs (the old Concept Model A and the WaterRower), these are related by a fixed constant. However, for variable damping ergs (Concept B,C, Rowperfect), the drag factor k can vary so the computer can re-evaluate these 'constants' stroke by stroke. So while the distance covered in each stroke is related to number of revs turned during that stroke, the factor may change from one stroke to the next if k changes.
The figure used for c is somewhat arbitrary - selected to indicate a 'realistic' boat speed for a given output power. Concept used to quote a figure c=2.8, which, for a 2:00 per 500m split (equivalent to u=500/120=4.17m/s) gives 203 Watts. However, Marinus van Holst ("Behind the Ergometer Display"), thinks the formula used is equivalent to
| (9.4) | P = 4.31 u2.75 |
giving 218 Watts for a 2:00 split (in my experience this is a bit high).
The Rowperfect definitely uses the cube law but the value c can be adjusted for different boat-types and rower's weight and sex.
Even if the damping factor k were to remain constant (Eq.9.1), there would not be a fixed relation between the average power for a piece and the mean split or speed. This arises from the non-linear relationship between power and speed (i.e. P = c u3 rather than P = c u), operating both from one stroke to the next and within individual strokes.
Take the case of rowing a 1000m piece in 4 minutes, either (a) at a uniform rate of 2:00/500m splits, or (b) rowing the first 500m at a steady 1:50 pace and the second half at a steady 2:10. Using Eq. (9.1) The average power for each of these two pieces will be
| (10.1) | (a): P = | c (500/120)3 = 72.34c |
| (10.2) | (b): P = | (110c/240).(500/110)3 + (130c/240).(500/130)3 = 73.86c |
Thus more power is required for the same average speed if the splits are uneven (this is also true for boats: see Basics (Section 5)).
However, even maintaining a constant split, there is not a fixed relationship between power and speed due to the variation of speed during the stroke itself. The split is derived from the mean fan rotational velocity ω through the stroke (Eq.9.2), while the power is proportional to an average of ω3. So, for example, rowing a lower rate will lead to a wider variation in ω throughout the stroke cycle, so that the average of ω3 will be larger, and more power is required to maintain the same average speed. (In boat terms, this is why 'sliding-rigger' boats were developed to reduce hull-speed variation during the stroke cycle - Basics, Section 5).
See section13 for the effect of rating on the actual power output of the rower.
Mechanical work W (a type of Energy) is defined as the average Power x time:
| (11.1) | W = P t |
If Power P is measured in Watts and time t in seconds, then the Work W is obtained in Joules. So, rowing a steady 200W for 30 minutes, you would generate an amount of mechanical work
| (11.2) | W = 200 x 1800 = 360 000 J = 360 kJ |
In physics, a 'calorie' is defined as the amount of heat energy required to raise the temperature of 1 gramme of water by 1 degree centigrade, giving 1 calorie = 4.2 Joules. Dieticians, on the other hand, use the term 'calories' differently - their 'calories' are 1000 times bigger ('kilo-calories', kC), so dividing 360 kJ by 4.2 gives the mechanical work done in terms of 'dietary calories': 85.6 kC
However, for the above workout you would actually get a displayed value approaching 500 kC, i.e. a factor 5 - 6 times larger. This is because the computer attempts to calculate the number of calories you burn up (effectively chemical energy contained in fats and carbohydrates) in order to generate the mechanical work. It uses the formula
| (11.3) | E = ( 4 W + 0.35 t ) / 4.2 [kC] |
where E is the displayed number of calories [kC], W is the mechanical work in kJ, calculated according to Eq. (11.1), t is the time in seconds. This assumes that the body actually requires 4 units of chemical energy to generate 1 unit of mechanical energy (i.e. 25% efficiency) plus a background consumption of 0.35 kJ/sec (=300 kC/hour).
Comment Jon Williams of Concept2 (12 Aug 04)
The 300 kC/hour has always been our best approximation for keeping alive and awake and going through the rowing motion at a reasonable stroke rate on an erg with the flywheel removed. This was arrived at from internal experiments and observations, data from Fritz Hagerman and studies done at Ball State.
For the above workout this would give
| (11.4) | E = ( 4.0 x 360 + 0.35 x 1800 ) / 4.2 = 493 [kC] |
This is a result of the action-reaction principle (Newton's 3rd Law). The force applied by your legs to the stretcher acts equally on you and the stretcher. In the static case, the stretcher is effectively attached to the whole planet so doesn't move - you do all the moving. In the dynamic case, the mass of the boat is much lighter (typically 10-20%) than you, so it moves further than you do.
This is not just a matter of the frame of reference: in the static (ergometer) you are actually performing more work in accelerating your body weight than in the dynamic (floating) case where the work is split between accelerating the boat and your body in opposite directions.
A 'dynamic' ergometer, such as the RowPerfect, attempts to simulate the reaction effect by having the stretcher/flywheel (together weighing approximately the same as a sculling boat) also mounted on a rail so that they also absorb most of the motion. Putting the Concept on 'slides' also simulates this effect, although since the Concept erg is much heavier than a sculling boat, it's not as realistic as the RowPerfect.
OK, those are the principles. Here's the maths ...
Imagine someone of mass Ma sitting in a boat of mass Mb, initially at rest then separating their centre of masses by a distance x in a time t with constant acceleration. Through conservation of momentum the final velocity Va, Vb (measured in opposite directions) of each component will be given by:
| (12.1) | Ma Va = Mb Vb |
Since the total separation s is achieved in a time t, at constant acceleration, then
| (12.2) | Va + Vb = s / t |
Then the total kinetic energy E in the final state (which has to be supplied by the rower) is given by
| (12.3) | E | = ½ Ma Va2 + ½ Mb Vb2 |
| = ½ Ma Va2 + ½ (Ma Va) ( s/t - Va ) | ||
| = ½ Ma Va (s/t) |
Putting in some typical numbers: Ma = 75 kg, t = 1 sec, s = 1 metre.
In the static case Mb is effectively infinite and Vb=0, giving Va = 1 m/s,
| (12.4) | Static: | E = 75 x 1 x 1 / 2 = 37.5 J |
In the dynamic case, Mb = Ma/5, so Vb = 5 Va, giving Va = 1/6 m/s and Vb = 5/6 m/s:
| (12.5) | Dynamic: | E = 75 x (1/6) x 1 / 2 = 6.25 J |
That is, on a static erg, the rower is required to put in six times as much energy accelerating/decelerating just their bodyweight, compared to a boat or a dynamic erg where the energy is split between the bodyweight and the boat/erg.
Cas Rekers (designer of the Rowperfect) has performed tests comparing the 'indicated' power output with and without the flywheel fixed - the subject gained about 10-20% power output in the second case, representing the additional power that could be applied to the flywheel instead of accelerating the bodyweight.
This is also one reason why the 'catch' on a static erg feels relatively 'slack' compared to a boat: the initial pressure on the feet is actually being used to decelerate/accelerate the body so the acceleration of the handle (as sensed by pressure in the hands) can only begin once the body has changed direction. The catch in the handle feels 'late' compared to the catch on the stretcher.
| (13.1) | t = 60 / R |
Assuming the rower moves up or down the slide a distance s, at the same constant speed in the stroke as well as the recovery, and changes direction instantaneously (!) at each end, the speed v will be
| (13.2) | v = 2 s / t = 2 s R / 60 |
For a rower of mass m, the Kinetic Energy associated with this motion is:
| (12.3) | U = ½ m v2 = 2 m ( s R / 60 )2 |
Assuming 'inelastic' ends of the stroke (i.e. no 'bounce' at the catch), then the rower, mass m, must supply sufficient work to recreate the kinetic energy U each time they change direction, twice every stroke, requiring a work rate (= power) of
| (13.4) | P = 2 U ( R / 60 ) = 4 m s2 ( R / 60 )3 |
Trying some numbers: for m=75 kg, s=1 m, R=30 str/min, this gives
| (13.5) | P = 4 x 75 x (1)2 x (30/60)3 = 37.5 W |
(Note that this is the same answer as Eq. 12.4, which is in fact derived for the same conditions). So the above rower is spending 37.5 W just moving up and down the slide at rate 30 (even without holding on to the handle).
Note that the rate appears as a cube term: changing the rate from 30 to 36 (a factor 1.2) results in a factor (1.2)3 = 1.7 increase in power loss. Similarly, rating 24 rather than 30 consumes (0.8)3 = 0.5 - only about half as much power is lost.
Also, note the appearance of the stroke-length term s2 - taller athletes are going to suffer a lot more at higher rates than shorter athletes.
There are a few more effects which have been ignored in all this which are probably smaller and also cancel out to some extent over the stroke cycle (ie some of the energy lost is regained in a different part of the stroke).
An approximate formula for the rate of change of air-pressure p with altitude is:
| (14.1) | p = p0 exp(-z/7000) |
where p0 is the sea-level pressure (approx 1000mb) and z is the altitude above sea-level in metres.
Self-calibrating ergs such as the Concept and RowPerfect would compensate for this by calculating a reduced drag factor (section 7), so still give an accurate measurement of the work done.
Eq. (14.1) corresponds to a pressure decrease of 1% for every 70m increase altitude. This means that, for a given lung-volume and breathing rate, the amount of oxygen taken into the bloodstream would also decrease by 1% for every 70m. If oxygen uptake through the lungs is the limiting factor in aerobic power output, then you would expect your erg power scores to fall off at the same rate (or split times to increase by 1% for every 210 m due to the cube relationship between power and speed, Eq. 4.5). E.g., in Denver (altitude 1500m), the air pressure is only 80% of the sea level value, so anyone moving up from sea level and trying a long-distance erg would probably find their power reduced by 20% (or times increased by approximately 7%).
Similarly, someone moving in the opposite direction (1500m down to sea level) would find 25% more oxygen in each lungful of air.
Notes