*No, not that kind of G-Force!*

I mean the extra "weight" we feel pressing us down onto the bike or into the track when riding at speed around the curved banking of a velodrome.

If you really want to read up on g forces, then this Wikipedia page covers it.

First of all, "g force" isn't really a force, rather it's a measure of a rate of acceleration. It's one of those slightly confusing expressions.

g force is a way to "normalise" accelerations relative to that we experience every day due to gravity on the surface of the Earth. It's a way to gauge how much we'd "weigh" when experiencing an acceleration that's more or less than 1g.

When we are standing on the ground, we experience 1g (units are usually expressed a "g", not to be confused with grams). We are not actually accelerating, since the ground is there to stop us from falling, so instead we feel a force we call weight.

While we normally use standard international unit of metres per second squared (m/s^2) to express accelerations, it's common to express accelerations relative to that we experience on the surface of the Earth, which is defined a 1g.

If you have a mass of 80kg and are standing on the ground then you'll weigh 1g x 80kg. That's why many confuse weight with mass. They are not really the same thing, mass is an intrinsic property of an object, weight is the force it pushes down on the ground with. It's just that on when sitting on the surface of the Earth, an object with a mass of 80kg will also weigh 80kg.

However if you are accelerating upwards away from the Earth's surface (imagine you're travelling upwards in a rapidly accelerating elevator or rocket), then you'll experience more than 1g.

How much more depends on the rate of acceleration. If you happens to be accelerating upwards at 9.81 metres per second per second (which equals 1g), then you'd experience 2g, being acceleration due to gravity + the extra acceleration of the elevator or rocket. If you were able to stand on some bathroom scales while that acceleration is happening, then you'll "weigh" 80kg x 2g or feel like you now weigh 160kg. ugh.

Now keep in mind that an acceleration can be a change in speed

e.g. when a travelling in a car that turns around a corner, even through the road speed of the car may not change, we experience what feels like a force pressing us towards the opposite side of the car. Such lateral accelerations can also be expressed relative to 1g. Modern Formula 1 racing cars for example are capable of generating lateral corning accelerations of up to 6g.

If you really want to read up on g forces, then this Wikipedia page covers it.

First of all, "g force" isn't really a force, rather it's a measure of a rate of acceleration. It's one of those slightly confusing expressions.

g force is a way to "normalise" accelerations relative to that we experience every day due to gravity on the surface of the Earth. It's a way to gauge how much we'd "weigh" when experiencing an acceleration that's more or less than 1g.

When we are standing on the ground, we experience 1g (units are usually expressed a "g", not to be confused with grams). We are not actually accelerating, since the ground is there to stop us from falling, so instead we feel a force we call weight.

While we normally use standard international unit of metres per second squared (m/s^2) to express accelerations, it's common to express accelerations relative to that we experience on the surface of the Earth, which is defined a 1g.

If you have a mass of 80kg and are standing on the ground then you'll weigh 1g x 80kg. That's why many confuse weight with mass. They are not really the same thing, mass is an intrinsic property of an object, weight is the force it pushes down on the ground with. It's just that on when sitting on the surface of the Earth, an object with a mass of 80kg will also weigh 80kg.

However if you are accelerating upwards away from the Earth's surface (imagine you're travelling upwards in a rapidly accelerating elevator or rocket), then you'll experience more than 1g.

How much more depends on the rate of acceleration. If you happens to be accelerating upwards at 9.81 metres per second per second (which equals 1g), then you'd experience 2g, being acceleration due to gravity + the extra acceleration of the elevator or rocket. If you were able to stand on some bathroom scales while that acceleration is happening, then you'll "weigh" 80kg x 2g or feel like you now weigh 160kg. ugh.

Now keep in mind that an acceleration can be a change in speed

__and/or__direction.e.g. when a travelling in a car that turns around a corner, even through the road speed of the car may not change, we experience what feels like a force pressing us towards the opposite side of the car. Such lateral accelerations can also be expressed relative to 1g. Modern Formula 1 racing cars for example are capable of generating lateral corning accelerations of up to 6g.

### g force on a velodrome

So whenever we are riding around the curved path on the turns of a velodrome, we are constantly accelerating towards the centre of the track, even though we may not be changing the bike's forward speed. As a result, we experience some lateral g forces when riding on velodrome, as well as the downward 1g due to gravity. What this means is the total g force we experience will be more than 1g. How much higher depends on our speed and the turn radius.

Calculating the rate of lateral acceleration when riding around a track is a pretty simple:

where:

- rider speed is in units of metres per second

- 9.81 is the rate of acceleration due to gravity in metres per second squared

- rider turn radius is measured in metres

### Estimating rider turn radius

To estimate a rider's turn radius, we start by estimating the track's turn radius.

First an overhead shot of a velodrome to get a sense of the general shape of a track (thanks to Google maps). This one happens to be a local outdoor track not far from where I live. It's a 333.33m concrete track. You can make out the faint blue band around the inside of the track, which is just inside of the track's black measurement line, the inner edge of which = 333.33 metres in length.

Now we can approximate the shape of the track as being two semi-circles joined by two straights. Here I superimposed some circles and lines to demonstrate:

Of course tracks are not exactly like this, in reality the shape of the turns are not perfectly circular, and the length of straights varies. They really do come in many different configurations. But as an approximation you can see from the diagram it's pretty good starting point. So to make a reasonable approximation of a track's turn radius at the black measurement line, all we need to know is the total length of the track, and the length of the straights.

Let's say you are riding a 250 metre track with 44 metre long straights. The turn radius around most of the turn will be approximately (250 - 2 x 44) / (2 x PI) = 25.8 metres.

Like I said, it won't be exactly that as in reality tracks have a variable turn radius but it's close enough for the purposes of this discussion. OK, that's great, we have the track's turn radius.

Let's say you are riding a 250 metre track with 44 metre long straights. The turn radius around most of the turn will be approximately (250 - 2 x 44) / (2 x PI) = 25.8 metres.

Like I said, it won't be exactly that as in reality tracks have a variable turn radius but it's close enough for the purposes of this discussion. OK, that's great, we have the track's turn radius.

Now back to our lateral g force formula:

Now this formula asks for the

__rider's__speed and turn radius, not the track's turn radius.The rider's speed and turn radius is based on the position of their centre of mass (COM). Since a rider leans over when riding around the banked turn of a velodrome, then their COM speed and turn radius is less than the wheel's speed and the turn radius at the track where the tyre is rolling along.

Another diagram to help explain. You might need to look at a larger version - so click or right click on it to view a larger pic. ('The pic of a leaning rider I found on this blog item. Hope they don't mind me borrowing it - I can't however vouch for the physics discussed in that item).

When a rider at speed rides around the turn of a banked velodrome, they are leaning over. That means the turn radius of their COM is less than the turn radius of the track where the tyre is rolling along. So if you want to calculate the g force on a rider you really should be using the COM turn radius and speed, which is a bit of a pain because bikes use speed sensors that measure the wheel's speed.

To calculate COM radius and speed, you then need to know the rider's COM height and their lean angle (from the vertical). It's a little basic trigonometry.

To calculate COM radius and speed, you then need to know the rider's COM height and their lean angle (from the vertical). It's a little basic trigonometry.

e.g. if a rider's COM is 1 metre (COM height will be about the same as floor to saddle height for a rider in an aggressive race position), and they are leaning over at 40 degrees from the vertical, then the rider turn radius = track turn radius - sin(40 degrees) x 1 metre. IOW we reduce the track turn radius by ~0.64 metres, which for our 250 metre track is about 2.5% of the track's turn radius. So not much, but enough that for some applications and analysis of track cycling data you need to take these things into account.

That then brings us to the question of how do we calculate the rider's lean angle? Well I'm going to leave that one for now because I'm getting further away from the issue of g forces than I'd really like. Suffice to say that we can reduce the estimated track turn radius and wheel speed by a couple of percent to estimate rider turn radius and rider COM speed.

### Are you still with me?

OK, so we can reasonably estimate the lateral g force of a rider travelling around the turns of a track. But of course the rider also feels the force of gravity pulling them downwards. So the total g force acceleration into the track is the sum of those two acceleration vectors:

Which can be expressed as follows:

So there you have it.

### How much g force does a rider feel in the turns?

If we use an estimated rider turn radius of ~25 metres for a typical 250 metre track with rider speeds ranging from 40km to 75km/h, here's what the estimated g forces are:

At elite hour record speeds of between 52-55 km/h, a rider will experience around 1.3g to 1.4g when riding the turns.

There is a physiological question to be considered, namely does this higher g force, which occurs twice per lap and lasts for about 2/3rds of the total time on track, cause any problems with blood circulation, is it sufficient to hamper performance?

I don't know, and I'm not sure if there's been anything more than speculation on this question. 1.3-1.4g doesn't sound like it'd cause too much of a problem to me, but who knows?

There is a physiological question to be considered, namely does this higher g force, which occurs twice per lap and lasts for about 2/3rds of the total time on track, cause any problems with blood circulation, is it sufficient to hamper performance?

I don't know, and I'm not sure if there's been anything more than speculation on this question. 1.3-1.4g doesn't sound like it'd cause too much of a problem to me, but who knows?

### Track sprinters

Things get much more interesting for the world's best track sprinters, who experience around 2g in the turns during their flying 200 metre time trial. If you have a 95kg track sprinter flying around at 75km/h, the bike, wheels and tyres are supporting the equivalent down force of over 200kg.

This is why track sprint bikes and wheels have to be made extra strong, and also partly why track sprinters run extra high pressure in their specialist tubular tyres.

This is why track sprint bikes and wheels have to be made extra strong, and also partly why track sprinters run extra high pressure in their specialist tubular tyres.

*Would you put a 200kg rider on your bike?*
## 3 comments:

Very nice! I especially like how you reduce the turning radius. But things get a bit more tricky if you consider that riders won't ride directly on the line. That gives back some of the shortcut taken by the COM.

In any case, this clearly has an influence on rolling resistance. If you're pulling 1.4 g's in turns, and turns are 60% of the track at the tire contact point trajectory, then that's a 24% increase in rolling resistance, approximately.

Crr variance is pretty complicated on a track.

The tyre is rarely normal to the track surface and so there is additional scrub going on (I don't know if that's the correct term), even on the straights the surface is banked (e.g. 12 degrees is common) and so the tyre is working to hold a line.

The question in turns is one of a sum of all the various factors. The COM turns a smaller radius, hence speed is reduced (e.g. by 2-3%) relative to the straights* and so air resistance is reduced accordingly (4-6%), while Crr goes up (e.g. 25%) due to the extra g force.

Since air resistance is >90% of the total resistance, then 4-6% x 90% still ~= 4-6% saving on air resistance, while 25% x <10% ~= 2% increase in rolling resistance.

* there is also a component of speed variability due to the changes in gravitational potential energy of COM between straights and turns. e.g say you are doing 15m/s with total mass of 90kg. KE = 10125J. Drop COM by 23cm via leaning over 40 degrees, and you gain 90kg x 9.8m/s^2 x 0.23m = 203J of potential energy, which is mostly converted to KE (although some losses due to increase in air resistance).

If all gravitational potential energy went to an increase in velocity then component of speed increase due to loss of gravitational potential is nearly 0.5km/h. That speed is then lost again when you exit the turn and COM regains height.

I meant to say wheel speed is increased in turns relative to straights, not COM is reduced relative to straights (since conservation of energy applies).

Post a Comment