## Calculations 2007-2011

The calculations presented in this page were performed before Jan 2012. Although I believe all are fairly accurate within the limits of the assumptions underlying them, some of them employ primitive assumptions necessary to permit simple solutions to Newton’s equations of motion using relatively simple algebraic methods. In January 2012, following detailed discussion with Robert Osfield recorded in the comments section of the page ‘Running: a dance with the devil’ I developed a computer program which provides a more realistic representation of the movement occurring during running, employing numerical integration of Newton’s equations allowing a realistic representation of the time course of vGRF.

Some calculations presented on this page, such as the calculation of vertical motion and airborne time (calculated on 29 Dec 07), the effect of gravitational torque (calculated on 6 Jan 2008); and the calculation of the increase in forward momentum of the foot and lower leg after lift-off from stance. (calculated, 13 Jan 2008) have been superseded by more realistic computations presented on the page ‘Calculations 2012 – present’

The calculations on this page are:

1) **Vertical motion and airborne time** (calculated on 29 Dec 07)

2) **The effect of gravitational torque **(calculated on 6 Jan 2008)

3) **The increase in forward momentum of the foot and lower leg after lift-off from stance**. (calculated, 13 Jan 2008)

4)** The role of gravitational torque in accelerated running.** (3 Jan, 2010)

**Vertical motion and airborne time**

(calculated on 29 Dec 07)

As described on the ‘Mechanics of Efficient Running’ page, vertical motion of the centre of gravity (COG) is undesirable because it increases impact forces at foot-strike and also requires energy to lift the body again, but freefall when airborne is inevitable. The free fall can be minimized by a high cadence, so that each airborne period is relatively short. This can be demonstrated by calculating the anticipated vertical motion for a given airborne duration.

The equation of uniformly accelerated motion is s=ut+1/2(at^2)

s is the distance travelled; u is initial velocity; t is the time taken

a is the acceleration due to gravity (=32 ft/sec/sec)

If we assume that there is an initial upwards velocity (mainly arising from imparted by elastic recoil at lift-off) just adequate to compensate for the fall during the airborne period, the value of s at the end on the airborne time is 0, giving u=-1/2(aT)

T is the airborne duration. (The minus sign for u indicates upwards motion.)

Therefore at time t during the airborne period, s=-1/2(aTt) +1/2(at^2)

The point of maximum vertical travel is the point where the rate of change of s with time is zero. Using simple calculus it can be shown that this occurs when aT=2at. Thus, as might be expected intuitively, the maximum vertical travel occurs midway through the airborne period at t=T/2

At t=T/2, the vertical distance traveled is given by s=-1/4(aT^2)

If T=200 millisec (0.2 sec) peak vertical distance traveled: s= 0.32 feet = 3.84 inches

If T=400 millisec (0.4 sec) peak vertical distance traveled: s = 1.28 feet = 15.36 inches

Thus, the total vertical travel in two airborne periods of 200 millisec is much less than the distance traveled in a single airborne time of 400 millisec.

As a vertical range of travel of 3.84 inches is tolerable, an airborne time up to 200 millisec might be acceptable, but longer times would lead to excessive free fall.

**The effect of gravitational torque**

(calculated on 6 Jan 2008)

When support is behind the Centre of Gravity (COG) gravity exerts a torque that causes rotational acceleration in a forwards direction (i.e. head moving forward relative to feet). This acceleration generates forwards angular momentum.

How large is the change in angular momentum due to gravitational torque?

Let us first perform the calculation for the situation where foot fall occurs under the COG, so that only forward angular momentum is generated. To make things simple, we will confine our calculations to the situation where the lean is small and there is no substantial bending at the hip, so that the stance leg and the trunk can be considered as a rigid object.

Let r=distance from COG to ground (approx 1 metre)

M = body mass

V= forward velocity (let us assume V=5 metres/sec)

g=acceleration due to gravity (=9.8 metres/sec/sec)

F = force producing torque = M.g.sin(theta)

theta = angle of lean of the body (from ankle to COG)

theta1 = lean at the point of lift-off from stance

L = angular momentum

dL/dt = rate in change of angular momentum

Torque = r.F= r.M.g.sin(theta)

Rate of change of angular momentum (dL/dt) = Torque = r.M.g.sin(theta)

Increase in angular momentum Delta(L) = intergral of dL/dt over the period on stance (in this instance while theta increases from 0 to theta1

Delta(L)=r.M.g.{integral[sin(theta)] d(theta)} / d(theta)/dt

[This integration with over the range of theta is only valid provided theta varies linearly with t, which is a valid assumption provided theta is small (i.e. substantially less than 1). See the discussion for a re-formation in terms of an integral over time.]

The integral of sin(theta) is cos(theta). For our purposes, we can calculate an approximate value of cos(theta) using the approximate formula:

cos(theta)=1-0.5theta.theta

Furthermore rate of change of theta is given by:

d(theta)/dt = V/r (where theta is in radians)

Substituting the value for the integral in the equation for Delta(L) gives:

Delta(L)= (M.r.r.g.theta1.theta1)/(2V)

The initial angular momentum (L) at foot-strike (due to inertia associated with forward velocity is r.M.V.

The proportional increase in angular momentum due to gravitational torque during the period on stance is given by:

Delta(L)/L = r.g.theta1.theta1/2V.V

Assume the body is on stance for a long enough time to allow the lean to increase by 0.1 radian (approximately 6 degrees)

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Substituting r=1 metre; g=9.8 metre/sec/sec; theta1=0.1 radian, V= 5 metre/sec gives:

Delta(L)/L = 0.002 = 0.2 per cent

That is: the increase in angular momentum due to gravitational torque while on stance is only 0.2 per cent of the initial angular momentum.

If the increase in lean is 0.15 radian (approx 10 degrees), the increase in angular momentum due to gravitational torque will be 0.4 per cent

This is only an approximate estimation due to the approximations made in the calculation, but is accurate enough for practical purposes.

**Conclusion**: gravitational torque produces a trivial contribution to angular momentum compared with that arising from the inertia associated with forward linear motion, under the circumstances dealt with in this calculation.

Note added 7th Jan 2008

As pointed out by Mike Stone in his comments (posted below), at a speed of 5 metres per sec, the body would rotate though 10 degrees in a period much less than the usual period on stance. In fact the equation that leads to that observation is the simple equation for the linear (tangential) speed of a rotating object, and does not depend on the assumptions made in my calculations. However, it does demonstrate that it would be virtually impossible to run at 5 metres/sec without extending the hip, allowing the leg to slope more than the trunk. In fact, this backwards bend at the waist can easily be seen in photos of athletes running at medium or fast speeds. Thus, if we want our model to provide a good description of running at speed, we need to treat the body as composed of at least two separate parts, joined by a hinge at the hip. This calculation is more complex, and I will defer it until I have a little more time.

3) **The increase in forward momentum of the foot and lower leg after lift-off from stance**. (calculated, 13 Jan 2008)

For the purpose of estimating the energy requirement as the leg swings forward after lift off from stance to the next foot-strike, it is useful to divide the swing into three distinguishable though partially overlapping segments. In practice, the smooth transitions between these distinguishable results in the foot following an approximately circular path below the torso. The three segments are:

1) Lift-off, in which the foot is lifted towards the hip. At the beginning of this segment, the foot is stationary on the ground. By the end of this segment it is moving forwards a little faster than the speed of the torso as it begins to overtake the torso. At the beginning of this segment the speed of each part of the lower leg (between ankle knee) will depend on how far from the ground it is. The bulky part of the calf is located approximately one third of the way from ground to centre of mass and will be moving forwards at approximately one third of the velocity of the torso, so for the purpose of the present calculation, we will assume that the lower leg is initially moving at approximately one third pf the velocity of the torso (A more precise calculation would require knowledge of the distribution of mass along the length of the lower leg, and an estimate of the initial velocity of each part of the lower leg.) During this segment of the swing, the foot and lower leg not only gain momentum but are also raised vertically. Thus, the foot and lower leg gain gravitational potential energy which must be provided by vertical forces (the vertical component of the pull and the vertical component of elastic recoil.) The thigh begins to swing forwards as the foot rises, but this should be an unconscious action and it is better to treat this swing as part of the next segment of action

2) The swing proper: in this segment, the thigh swings forward from a hinge at the hip, while the lower leg swings forward from the knee. As a result of these two simultaneous swinging actions, the foot moves rapidly forwards and overtakes the torso. At the end of this segment the knee should still be moderately flexed. When running at a relaxed speed, these swinging actions at hip and knee should not be driven consciously, as this would create a risk of over-striding. If the swing is actively driven it is possible to run faster, but then active braking of the leg by means of active hip extension would be required during the final segment, to avoid over-striding. This forceful action would increase energy consumption and risk of injury.

3) The foot fall: the whole leg swings back in a pendulum-like action from the hinge at the hip, to bring the foot beneath (or perhaps slightly in front of ) the torso at the point of foot strike. The tension of the muscles of the thigh and calf must be adjusted to prevent collapse at the knee and ankle on foot-strike. In this phase, the major energy consumption is likely to be the energy required for isometric tensioning of the thigh and calf muscles. However, this should be left to unconscious brain mechanisms. If the foot is landing too heavily, simply think of decreasing the sound of the thud and the brain will make the required adjustments of muscle tension.

Thus, the main energy consuming actions required during the swing phase are:

1) forward acceleration of foot and lower leg immediately after lift off.

2) vertical lifting of foot and lower leg

3) forward swing of thigh in the swing proper. This should be as relaxed as possible.

4) tensioning of the thigh and calf muscle during foot fall

In this calculation we will estimate the amount of momentum that must be gained by the foot and lower leg during lift off, and compare this to the forward impulse provided by the forward directed horizontal component of ground reaction force (GRF) during preparation for lift-off in the final 100 milliseconds of the stance phase.

This computation requires an understanding of the impulse – momentum change theorem, which can be derived from Newton’s second law of motion. When a force F acts on an object for a time t, the momentum of the object increases. The increase in momentum is proportional to the strength of the force and also to the length of time the force is applied. Thus, the increase in momentum is equal to the product of force by time (which is known as the impulse).

Impulse = F * t

Increase of momentum = mass * increase of velocity

F * t = mass * change in velocity

(see http://www.physicsclassroom.com/Class/momentum/U4L1c.html)

We will perform the calculation for a 70 Kg runner running with a steady velocity of 4.5 metres per sec. We will assume that the foot weighs 1 Kg and the lower leg from ankle to knee weighs 3.5 Kg

*The gain in momentum*

Momentum acquired by foot = m * .(v1-v0)

m= mass of foot = 1 Kg

v0= initial velocity of foot = 0

v1= final velocity of foot = 4.5 (approximately)

Momentum acquired by foot= 4.5 Kg metre/sec (approximately)

Momentum acquired by lower leg = M * (V1-V0)

M= mass of lower leg = 3.5 Kg

V0=initial velocity of lower leg = 1 .1 metre/sec

V1=final velocity of lower leg= 4.5 metre/sec (approximately)

Momentum acquired by lower leg = 3.5×3.4= 11.9 Kg metre/sec

Total momentum acquired by lower leg and foot during lift-off = 15.4 Kg metre/sec.

*The forwards impulse from horizontal GRF *

The forwards impulse delivered to the foot via the horizontal component of ground reaction force (GRF) generated by the oblique push of the leg on the ground during the final 100 milliseconds on stance can be computed from values of GRF measured using a force plate. Cavanagh and la Fortune (Journal of Biomechanics, 13, 397-406, 1980) measured the ground reaction forces in runners landing on the mid-foot (and also in heel strikers) running at 4.5 metre/sec. For those landing on mid-foot, they found that the horizontal component of GRF during the final 100 milliseconds on stance was directed forwards (as expected from the oblique push of the leg on the ground during this time), and exhibited a shallow peak rising to a maximum value of 0.45 times body weight around 50 milliseconds before the completion of lift-off. The average value of the horizontal GRF during the final 100 milliseconds was approximately 0.2 times body weight

Weight of runner = 70 Kg

Mean F= 0.2 BW = 14 Kg wt = 137 newtons

Duration =0.1 seconds

Impulse = F * t = 13.7 Kg metre/sec

Thus the impulse imparted by the forward directed component of GRF provides approximately 90% of the gain in momentum of the foot and lower leg during lift-off. The remainder must come from other sources that exert a forward propulsive force on the flower leg and foot. The major source is likely to be contraction of the hamstring muscle.

The main simplifying assumptions made in this computation are:

1) the lower leg (ankle to knee) has been treated as a single mass located at the level of mid-calf. A more precise calculation would take account of the distribution of mass along the lower leg.

2) the forward velocity of foot and calf at the end of the lift off has been taken to be equal to the velocity of the torso. In fact, by this stage, the foot is likely to be moving a little faster than the torso, though perhaps the extra increment of velocity can be regarded as a result of the forward swing of the thigh and any energy required for this relaxed swing can be regarded as part of the energy cost of the second segment of the swing.

**4) The role of gravitational torque in accelerated running** (3 jan 2010)

In the discussion on my blog on 31 Dec 09, Simbil encouraged me to look more closely at the paper by Fletcher, Dunn and Romanov describing the observations of accelerated running that led them to suggest that gravitational torque plays an important role in acceleration. The paper can be accessed at:

http://w4.ub.uni-konstanz.de/cpa/article/viewFile/3291/3092

The paper is frustrating because of apparent inconsistencies and peculiar mathematics. Here are the problems that must be addressed in making sense of that paper.

First of all, one of the main points the authors make is that the maximum horizontal acceleration of the COM occurs before the maximum horizontal GRF. However equation 1a shows that horizontal acceleration of COM is strictly proportional to horizontal GRF. Therefore, it appears that either figure 1 and the main conclusion of the paper are wrong, or equation 1a is wrong.

Secondly, even if we accept that figure 1 is correct, it appears odd to conclude that gravitational torque might cause acceleration on the grounds that horizontal GRF reaches its peak value after the point of maximum acceleration of the COM while the data also show that gravitational torque reaches its maximum even later than horizontal GRF. In fact one does not even need to look at the data; gravitational torque must be greatest at the end of stance around the time when the horizontal distance between COM and point of support is at its maximum.

Thirdly, the authors claim that equation 5 shows gravity’s effect on GRF. The logic for this statement is not clear. The derivation of equation 5 appears very peculiar.

The authors state that equation 5 was derived by substitution of equations 3 and 4 into equation 2. In fact this substitution leads to the equation:

dω/dt = (g/kr)sinθ where k is a parameter determined by the shape of the body. Anyone with a mathematical inclination can demonstrate that k is (rho/r)*(rho/r) where rho is the radius of gyration about an axis through the point of support, and r is the distance between COM and point of support. If the body has a long, thin pencil shape, k is 4/3).

The authors leap from this to equation 5: dω/dt = ω^{2}sinθ. If equations 2,3,4 and 5 are correct, ω must be sqrt(g/kr). At the point where COM passes over the point of support (ie, when θ=0), horizontal velocity is given by r*ω. Therefore, equation 5 implies horizontal velocity at this point is r*sqrt(g/kr). Thus, equation 5 predicts that horizontal velocity at mid-stance is a constant for a runner of a particular height and shape. For a pencil-shaped runner whose COM is 1 metre off the ground at mid-stance, the predicted velocity would be 2.71 m/sec. It is implausible to propose that all runners of the same shape and height reach the same speed on the second stride. It can be seen from table 2 that for the HT runner the horizontal velocity of the COM is around 3.4 m/sec, while for the Pose runner it is between 4.2 and 4.5 m/sec, when the COM passes over the point of support, yet the two runners have similar height and weight.

One possible problem in the derivation of equation 5 might be that equation 4 is incorrect. This equation appears to assume that r is constant. This might not be a valid assumption, as it would lead to the COM rising at mid-stance whereas I believe that in practice, the opposite is the case.

I believe that the conclusions of Fletcher, Dunn and Romanov should be regarded with skepticism until these inconsistencies and peculiarities have been sorted out.

January 6, 2008 at 5:55 pm |

Comment on the computation of gravitational torque

Canute – I applaud your willingness to get your hands dirty and do your best to get a proper handle on the forces which act upon us when we’re running.

Only problem is, in this case, the Maths is wonky.

That line

Delta(L)=r.M.g.{integral[sin(theta)] d(theta)} / d(theta)/dt

doesn’t make sense – I think you’re trying to invoke the chain rule somewhere along the way, but it doesn’t work like that.

correct would be

Delta(L)=rMg{integral[(dL/d(theta)).(d(theta)/dt) dt ]}

which unfortunately isn’t an expression you can conveniently integrate.

The way to approach it is to use the angular equivalent of Newton’s third law:

I(theta)” = rMgsin(theta)

where I is the moment of inertia of the rotating object, and the ‘ denotes a derivative w.r.t. time (dot over the top would be more appropriate, but the ‘ is easier to type)

This does lend itself to further manipulation:

Iw.dw/d(theta) = rMgsin(theta)

where w = (theta)’ (chain rule)

So integral { Iw dw } = integral { rMgsin(theta) d(theta) }

which are both easily integrated expressions (but pay attention to your initial conditions to get the constants of integration right) – in fact what you then get is an equation expressing the exchange of kinetic and potential energies of the system.

Not sure what shape you’re assuming our falling human to be – you could idealize him/her as a rod of length 2r – in which case I = (4Mr**2)/3

( r**2 here denotes r squared)

Unfortunately if you then try to integrate again to get an explicit function for theta in terms of time, you get a nasty elliptic integral to deal with.

I’ve just returned to studying applied Maths – I’m about to start the second year of a MSc course with OU . I’m fascinated by the idea of using my subject to model human motion (running in particular) – but it’s an incredibly complex model!

What did you make of that comment a little while ago about Michael Johnson spending more time in the air than any other runner? It made me think a bit, in terms of what that implies when he’s in contact with the ground. I may comment on that in a little while.

best wishes

Mike S

January 6, 2008 at 6:55 pm |

Thanks Mike,

I am dredging up mathematics and physics from over 35 years ago, so I am delighted that a person with current mathematical expertise is helping.

However, I think if there is an error, it might be in my physics of rotational motion rather than in the mathematics.

My intuitive use of the chain rule might have looked a bit peculiar, but as you point out the real issue is the correct equation to describe change on angular momentum

Based on a distant memory, I was starting for the premise that rate of change in momentum is equal to torque, and that torque is r.F where F is the tangential force acting at distance r from the axis of rotation.

Starting once again from this premise, let me reframe the problem:

dL/dt=r.F

F=M.g.sin(theta)=M.g.theta (provided theta <<1)

theta=V.t/r (for theta<<1)

Therefore: dL/dt=M.g.V.t

Integrating from t=0 to t=t1 (time of lift off from stance) gives:

Delta(L)=0.5M.g.V.t1.t1

t1= theta1.r/V (for theta1<<1)

Delta(L)=M.g.r.r.theta1.theta1/2V

L=r.M.V

Delta(L)/L=r.g.theta1.theta1/2V.V

This is identical to the expression that I had obtained with my intuitive application of the chain rule. So it seems to me that provided my initial premise (dL/dt=r.F) is correct, my calculation is OK.

However, in view of the rustiness of my memory of rotational mechanics, I would be very grateful for your opinion on my premise (dL/dt=r.F)

January 6, 2008 at 7:38 pm |

Mike,

I have just checked torque in Wikipedia and it seems that Wikipedia confirms my distant memories of rotational mechanics.

Have a look at:

http://en.wikipedia.org/wiki/Torque

So my premise looks OK. Therefore, unless you can spot a problem with my algebra, I think my original calculation must be correct

Canute

January 6, 2008 at 8:49 pm |

Mike,

With regard to my application of the chain rule in the integration, I think it is valid provided theta varies linearly with time which is the case for small rotations. I have added a note to my original calculation pointing this out. Provided no-one finds any other problems with my calculation, I will edit the calculation, presenting the integral as an integral over time rather than theta, as I think this will be easier for people to follow.

With regard to your comment about Michael Johnson, the first issue is to distinguish between the length of time he is airborne and the distance for which he is airborne. Because he is very fast, he will certainly go a lot further in a given period of time than I would. It is the length of time airborne that matters with regard to waste of energy through freefall.

If in fact he does actually spend more time in the air than me (or even Haile G), then he will waste more energy falling. However, a runner can afford to do this over a short distance – waste of glycogen is irrelevant in a sprint, but matters hugely in a marathon.

If he is airborne for longer, his range of vertical motion will be higher than you would find in an elite marathon runner. Maybe you could check this by comparing him with Haile G.

January 6, 2008 at 9:20 pm |

Canute

Now you’re making me think!

no problem with the premise – that’s absolutely fine

But, looking thru your whole derivation, while I have no real problems with the algebraic manipulation, I have issues with other premises – the ones on which some of your approximations are based.

In order of increasing significance:

1) I’m a little bit uneasy about the sin(theta) ~= theta assumption here. Fine for a simple pendulum when the effect of the forces acting ensure that theta remains small – but here they don’t. OK – let’s assume that throughout the whole motion (i.e. up to the point where the foot leaves support), theta is sufficiently small that we can neglect terms of order (theta cubed).

2) Your expression for the angular momentum of the system is only valid if the body consists of a single point of mass M located at r . You need to take account of the fact that the mass of the body is spread out around the axis of rotation – i.e. you need to express this in terms of moments of inertia.

3) The assumption that d(theta)/dt = v/r – now this REALLY does cause me problems.

Firstly, there’s the assumption that a generalised co-ordinate (theta) whose second time derivative must be non-zero (because we know the angular momentum increases) has a constant first derivative.

Secondly, consider it from the real-world physical point of view. If we are to take your example of a time on support sufficient for theta to increase by 0.1 radians, well…. v/r = 5 radians per second, giving us a time on support of 0.02 seconds….even Michael Johnson’s going to struggle to manage that.

Thirdly – the centre of mass is travelling at a mean velocity of 5m/s – but so’s any other part of the body . So why can’t we base our approximation for angular velocity on the average velocity of the top of the head ( i.e. d(theta)/dt = v/2r ) ?

Sorry, but I do think there are problems with this derivation.

best regards

Mike S

January 6, 2008 at 10:08 pm |

Mike,

I should emphasize that my calculation does assume that the body is fairly rigid, with a constant alignment of ankle, hip and shoulder in a near straight line. That is what Pose recommends. Under this circumstance, I think all my assumptions are fairly reasonable when the lean is in the range 0 to 0.15 radians (0 to approx 10 degrees) which I think covers the Pose range.

It should also be noted that I am neglecting the other (swinging) leg on the grounds that it is much lighter than the torso and therefore makes a relatively minor contribution. There is no doubt that my model is only an approximation, but I would argue that it is good enough to give a practically useful estimate of the effect of gravitational torque.

Best wishes

Canute

January 6, 2008 at 10:18 pm |

Mike,

Finally, with regard to your final point about the time to rotate through 0.1 radians, if I have employed an over-estimate of the amount of rotation compared with that which would occur in the real world, then my calculation is too conservative. If you make theta1 smaller, then the effect of gravitational torque is even less than my calculation. However, if the person starts with a substantial initial lean, then the integration should be over a range that starts from a non-zero value. However, provided the final lean does not exceed about 0.15 radians, I think my calculation provides a reasonable estimate of the upper limit to the contribution from gravitational torque.

January 6, 2008 at 10:48 pm |

Canute,

Bit confused by your last post. If you have indeed overestimated the amount of rotation (quite possible), then the time on support is even less than 0.02s, so it’s an even less realistic figure. My understanding is that even for an elite athlete, a support time as short as 0.1s would be an amazing feat. So how does an even less realistic estimate for one of your parameters make your conclusion more plausible?

One of Richard Feynman’s more brutal quotes springs to mind:

“It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong”

sorry

Mike S

January 6, 2008 at 11:04 pm |

Mike,

On further thought about the assumption that ankle, hip and shoulder remain in a straight line (which is one of the rules for good Pose running originally specified by Dr Romanov), there is no doubt that in practice runners disobey this rule once the pace has increased beyond a fast jog. What actually happens is that the hip extends while the runner is on stance so that the leg on stance rotates through a larger angle than the torso. I accept that once we allow violation of the rules of Pose, it would be better to employ a model in which the body consists of two elements joined by a hinge at the hips. However, I think my model will still give a fairly good answer if we regard theta as the angle between the line from ankle to a notional dynamic COG that moves a little as the body bends. I would still ague that provided this dynamic COG does not move far, the assumptions of my model are OK. However, maybe you could try solving the problem by numerical integration for a model in which the body is allowed to hinge at the hip.

January 7, 2008 at 1:01 am |

Mike.

Sorry, I didn’t make myself clear enough in my response at 10:18 and I had not seen your comment at 10:48 when I pressed submit on my comment at 11:04 so the logic has got a bit out of order.

The point about the hinge at the hip is that the leg does move through a larger angle than the torso; the torso very rarely goes beyond about 6-10 degrees (except in sprinters) but even in long distance runners, the leg gets left behind due to hip extension and may move through a larger angle (even though that is counter to the original rules of Pose). So I think that to really satisfy ourselves we need to do the calculation assuming a hinge at the hip.

I would still anticipate that provided the angle from the ankle to the COG at point of lift off does not exceed 10 degrees, my calculation will not be too far out, but I think the calculation need to be done with the hinged model. As the final angle of the leg might exceed 10 degrees, I think we will need numerical integration.

Canute

January 7, 2008 at 9:56 am |

Great stuff Canute. A couple of comments regarding some of the assumptions – the first is the body alignment in Pose. The alignment is BOF – Hip – Shoulder when in stance phase (the Pose), the Hip – Shoulder alignment remains after stance but the BOF – Hip alignment separates as the leg is left behind. So there are 2 rotations about the hips in Pose. The second is the lean angle, 22 degrees has been banded about on the Pose site as a max angle of lean for elite runners, I’ve no idea where it came from originally though I’m afraid. Good luck if you attempt calculations on a double rotating system, I watch with interest.

January 7, 2008 at 10:42 pm |

Thanks Simon. I accept that the alignment of BOF -hip-shoulder is a better description of Pose. The relevant assumption in my calculation is that the body does not bend appreciably at the hip. In these circumstances, the relationship of BOF-hip-shoulder and also ankle-hip -shoulder will remain constant.

Mike’s observation of the unrealistically short time on stance at 5 metres/sec (which is actually a consequence of the simple equation for the speed of a rotating object and does not depend on the assumptions of my model) demonstrates that at medium or high speeds, the hip must extend. This is obvious from photos of runners, but nonetheless has caused some confusion for people who attempt to apply Dr Romanov’s recommendation about the alignment of BOF, hip and shoulder in Pose in too simplistic a manner. There was a somewhat confused discussion about this point on the PoseTech forum recently.

I accept that some runners might achieve a lean of 22 degrees. To deal with angles as large as this, I think the integration over time (or theta) will have to be done numerically. I will have go at this computation when I have the time