Cycling training

[DaveB]

On the flat I can average 15mph. The hills are the problem. This is all on a big heavy mountain bike with big knobbly tyres, at some point I will switch to a road bike.

Get some road tyres on it now, will make a big difference and switch to a proper road bike as soon as you can if thats what you intend to do the L2B on. Riding positions are very different and you'll need to get used to it.

No reason you can't do the L2B on the mountain bike but deffo ditch the knobbly tyres. Depending on the gearing the MTB could actually be easier on the hills, they are designed to go up the steep stuff at low speeds. You will end up at the back of the pack on an MTB though, nothing like as quick as a road bike.

[doodab]

No. Rolling friction is directly proportional to weight.

Ignoring Aero drag, which as you say will not dominate until above about 14mph, the formula for power in cycling is:

g * Mass * Velocity * (Frictional co-eff + hill grade)

You are of course correct, my point was that on the flat at a steady pace you will not actually notice the extra weight. Unless of course you are plodding.

[TimberWolf]

You are of course correct, my point was that on the flat at a steady pace you will not actually notice the extra weight. Unless of course you are plodding.

I've got some old javascript knocking around into which I've plugged in some numbers to see the difference numerically rather than getting my pen out and doing it algebraically. This includes aero drag, but there's no hill or wind, in this case.

Comparing a 100kg person cycling at 6 m/s (14mph) versus a slimmed down 90kg person (both inc. bike) doing the same speed. Knobbly tyres and no wind nor hill. Rolling drag. co-eff = 0.0120 (knobbly tyres; narrow tubular tyres can be 0.00475), aero. drag co-eff = 0.185 [ (1/2 * 1.3 kg/m^3 * 0.7 * 0.4m^2) ]

1) 100kg person:

Speed = 6 m/s = 22 km/h = 14 mph (no wind)
Rolling drag = 71 watts (64%) Aero drag = 40 watts (36%)

Power req'd = 111 watts (0.1 HP) (*4)
Force req'd = 12N + 7N = 18N

Energy req'd to travel 100 km = 1842 KJ = 0.5 kWh (*4)
[ car = 83 kWh, 747 = 42kWh per passenger per 100 km (level flight) ]

MPG = 5533 (/4) = 1383 MPG

Will roll to stop in 121 m in 44 secs
Drag as a function of weight (D/W) : 0.019 (747 = 0.07, ocean birds = 0.1)

2) 90 kg person:

Speed = 6 m/s = 22 km/h = 14 mph (no wind)
Rolling drag = 64 watts (61%) Aero drag = 40 watts (39%)

Power req'd = 103 watts (0.1 HP) (*4)
Force req'd = 11N + 7N = 17N

Energy req'd to travel 100 km = 1724 KJ = 0.5 kWh (*4)
[ car = 83 kWh, 747 = 42kWh per passenger per 100 km (level flight) ]

MPG = 5911 (/4) = 1478 MPG

Will roll to stop in 119 m in 43 secs
Drag as a function of weight (D/W) : 0.02 (747 = 0.07, ocean birds = 0.1)

So 10 kg in this case made 8 watts difference (111 versus 103 watts). I didn't display enough significant figures to be able to compare the drag forces for such small differences. The *4 is a reminder to me that humans are only about 25% efficient, so the calorie count would be 4 times higher.

[suityou01]

I've got some old javascript knocking around into which I've plugged in some numbers to see the difference numerically rather than getting my pen out and doing it algebraically. This includes aero drag, but there's no hill or wind, in this case.

Comparing a 100kg person cycling at 6 m/s (14mph) versus a slimmed down 90kg person (both inc. bike) doing the same speed. Knobbly tyres and no wind nor hill. Rolling drag. co-eff = 0.0120 (knobbly tyres; narrow tubular tyres can be 0.00475), aero. drag co-eff = 0.185 [ (1/2 * 1.3 kg/m^3 * 0.7 * 0.4m^2) ]

1) 100kg person:

Speed = 6 m/s = 22 km/h = 14 mph (no wind)
Rolling drag = 71 watts (64%) Aero drag = 40 watts (36%)

Power req'd = 111 watts (0.1 HP) (*4)
Force req'd = 12N + 7N = 18N

Energy req'd to travel 100 km = 1842 KJ = 0.5 kWh (*4)
[ car = 83 kWh, 747 = 42kWh per passenger per 100 km (level flight) ]

MPG = 5533 (/4) = 1383 MPG

Will roll to stop in 121 m in 44 secs
Drag as a function of weight (D/W) : 0.019 (747 = 0.07, ocean birds = 0.1)

2) 90 kg person:

Speed = 6 m/s = 22 km/h = 14 mph (no wind)
Rolling drag = 64 watts (61%) Aero drag = 40 watts (39%)

Power req'd = 103 watts (0.1 HP) (*4)
Force req'd = 11N + 7N = 17N

Energy req'd to travel 100 km = 1724 KJ = 0.5 kWh (*4)
[ car = 83 kWh, 747 = 42kWh per passenger per 100 km (level flight) ]

MPG = 5911 (/4) = 1478 MPG

Will roll to stop in 119 m in 43 secs
Drag as a function of weight (D/W) : 0.02 (747 = 0.07, ocean birds = 0.1)

So 10 kg in this case made 8 watts difference (111 versus 103 watts). I didn't display enough significant figures to be able to compare the drag forces for such small differences. The *4 is a reminder to me that humans are only about 25% efficient, so the calorie count would be 4 times higher.

So being a fatty means more watts which means more calories spent so quicker weight loss?

Brill!

[TimberWolf]

So being a fatty means more watts which means more calories spent so quicker weight loss?

Brill!

Getting rid of those knobbly tyres and the tyre around your waist would improve your efficiency considerably. You've got to be fit to be fat on a bike.

[Churchill]

I've got some old javascript knocking around into which I've plugged in some numbers to see the difference numerically rather than getting my pen out and doing it algebraically. This includes aero drag, but there's no hill or wind, in this case.

Comparing a 100kg person cycling at 6 m/s (14mph) versus a slimmed down 90kg person (both inc. bike) doing the same speed. Knobbly tyres and no wind nor hill. Rolling drag. co-eff = 0.0120 (knobbly tyres; narrow tubular tyres can be 0.00475), aero. drag co-eff = 0.185 [ (1/2 * 1.3 kg/m^3 * 0.7 * 0.4m^2) ]

1) 100kg person:

Speed = 6 m/s = 22 km/h = 14 mph (no wind)
Rolling drag = 71 watts (64%) Aero drag = 40 watts (36%)

Power req'd = 111 watts (0.1 HP) (*4)
Force req'd = 12N + 7N = 18N

Energy req'd to travel 100 km = 1842 KJ = 0.5 kWh (*4)
[ car = 83 kWh, 747 = 42kWh per passenger per 100 km (level flight) ]

MPG = 5533 (/4) = 1383 MPG

Will roll to stop in 121 m in 44 secs
Drag as a function of weight (D/W) : 0.019 (747 = 0.07, ocean birds = 0.1)

2) 90 kg person:

Speed = 6 m/s = 22 km/h = 14 mph (no wind)
Rolling drag = 64 watts (61%) Aero drag = 40 watts (39%)

Power req'd = 103 watts (0.1 HP) (*4)
Force req'd = 11N + 7N = 17N

Energy req'd to travel 100 km = 1724 KJ = 0.5 kWh (*4)
[ car = 83 kWh, 747 = 42kWh per passenger per 100 km (level flight) ]

MPG = 5911 (/4) = 1478 MPG

Will roll to stop in 119 m in 43 secs
Drag as a function of weight (D/W) : 0.02 (747 = 0.07, ocean birds = 0.1)

So 10 kg in this case made 8 watts difference (111 versus 103 watts). I didn't display enough significant figures to be able to compare the drag forces for such small differences. The *4 is a reminder to me that humans are only about 25% efficient, so the calorie count would be 4 times higher.

When I went to school 11 + 7 = 18

Force req'd = 11N + 7N = 17N?????


And 12 + 7 = 19

Force req'd = 12N + 7N = 18N

Shirley?

[TimberWolf]

When I went to school 11 + 7 = 18

Force req'd = 11N + 7N = 17N?????


And 12 + 7 = 19

Force req'd = 12N + 7N = 18N

Shirley?

The numbers are rounded, what I said. If you really want to see more significant figures, I'll update my script.

[Churchill]

The numbers are rounded, what I said. If you really want to see more significant figures, I'll update my script.

Your figures just looked weird. Don't throw your toys out of the pram.

[TimberWolf]

Your figures just looked weird. Don't throw your toys out of the pram.

Have you never heard the expression 2 + 2 = 5 for sufficiently large values of 2?

[Churchill]

Have you never heard the expression 2 + 2 = 5 for sufficiently large values of 2?

assuming that your "2" is less than "2.5" yes.

However, considering you're gaining 10% with your rounding in that case why just stick to integers?