Is this climbing/gearing calculation correct?

[Twilkes]

Assuming no air resistance, if I'm riding up a 14% climb at 90rpm in 34-32 gear, if I then ride up a 7% climb in 34-16 gear at 90rpm, is that theoretically the same power output on both climbs? 50% reduction in gradient, 50% reduction in sprocket teeth.

if so it would make training for a climbing day easier as the 7% climb is much closer to me.

 

[All uphill]

Same torque required.
Less power.

I think.

Edit: second thoughts. In case 2 you will be going at twice the speed so, yes, the same power output per minute.

I think

 

[ColinJ]

You must weigh an awful lot less than me, or be incredibly fit...

I plugged those numbers into a couple of online calculators and I would have to be averaging well over 500 Watts to go at those speeds on those climbs!

I would be doing nearer to 40 rpm.

 

[T4tomo]

You must weigh an awful lot less than me, or be incredibly fit...

or exaggerating his /her ability

 

[ColinJ]

But yes - charging up the 7% climb at twice the speed of the 14% climb would need (near enough) the same power.

 

[Dogtrousers]

Well my back of fag packet calculations say that those two gearings at 90rpm will give roughly 12 and 24 km/h respectively on 622mm rims with 25mm tyres. Depends on your tyres to get it exactly. But what's important is that 24 is twice 12

I went to http://www.bikecalculator.com/ and left all the variables at their default values and asked for the power required for:
24 km/h up a 7% climb and it said 408W 478W (Edited because I misread it first time)
12 km/h up a 14% climb and it said 403W

So yeah. Pretty much the same. Similar but drag wll be a bit more significant at 24 km/h I guess ... to the tune of 75W.

I put my own weight and the weight of my bike into it and pressed calculate and it just laughed at me.

 

[si_c]

Same torque required.
Less power.

I think.

Edit: second thoughts. In case 2 you will be going at twice the speed so, yes, the same power output per minute.

I think

Power is effectively Torque x RPM so assuming same torque output from the rider then yes the power output would be the same.

 

[si_c]

Power is effectively Torque x RPM so assuming same torque output from the rider then yes the power output would be the same.

Just to add, speed would not be double as aerodynamic drag increases with the square of the speed, so proportionally more power is used to overcome air resistance at the higher speed meaning less is available for overcoming gravity.

 

[Supersuperleeds]

14% or 7%, either I'd be blowing out of my rear (and I don't mean tyre)

 

[ColinJ]

24 km/h up a 7% climb and it said 408W
12 km/h up a 14% climb and it said 403W

So yeah. Pretty much the same. Drag wll be a bit more significant at 24 km/h I guess

I have MY reading glasses on, and to me it looks like 478 W at 24 km/h up 7%!

But yes - charging up the 7% climb at twice the speed of the 14% climb would need (near enough) the same power.

Just to add, speed would not be double as aerodynamic drag increases with the square of the speed, so proportionally more power is used to overcome air resistance at the higher speed meaning less is available for overcoming gravity.

I didn't think that the air resistance would be particularly significant between 12 and 24 km/hr but I just plugged the numbers into bike calculator and was surprised by how significant it actually was...

The 403 W that gave 12 km/h up 14% only gave 21 km/h up 7%!

So, when we see the pros drafting on moderately steep climbs they actually ARE getting a significant benefit from doing so.

 

[Dogtrousers]

I have MY reading glasses on, and to me it looks like 478 W at 24 km/h up 7%!

Ooops. Can I borrow your glasses? I'll edit my post.

So it looks like the drag contribution is a bit more than I thought. 75W.
But the OP did say "assuming no air resistance"

 

[si_c]

So, when we see the pros drafting on moderately steep climbs they actually ARE getting a significant benefit from doing so.

Well, i'm not sure what counts as significant but there will definitely be a benefit. On the flat at around 12km/h drag is from rolling resistance is roughly equivalent to aerodynamic drag, but at 24km/h that theoretically changes to about 80% from aerodynamic drag (assuming identical position on the bike and no wind at all). So given how fast pro cyclists go up a hill they might save as much as 30% of that aerodynamic drag, which is not nothing, but on steeper gradients this will likely fall into a handful of watts.

 

[Sallar55]

Going twice as fast against gravity? I would suspect that you are at it. Level road twice the speed , you air resistance increases dramatically so that's another fail.

 

[ColinJ]

So given how fast pro cyclists go up a hill they might save as much as 30% of that aerodynamic drag, which is not nothing, but on steeper gradients this will likely fall into a handful of watts.

Yes, so when their teammates drag them up the 7% as fast as they can they are doing some good. Their leader leaves them behind when they can't do it any longer and tackles the following 12%/15%/whatever-high-% solo.

I used to think that the main benefit was psychological, but obviously it is not.

 

[Twilkes]

or exaggerating his /her ability

Hey, I said 'if'.

Yeah I'm not getting up a 7% climb at 24kmh, best one was 18kmh could probably do it 19kmh now at a push, so closer to 70rpm, usual cadence is 85-90rpm. 105kg and 10kg bike so still around 450w but it's only about 3.5 minutes long and FTP is just over 300w so it's not too much of a stretch.

It's more the principle that as long as I don't shift down a gear it'll be simulating a much steeper climb, as there are only a few round here and they don't tend to last very long. Doing a big round trip up north at the end of July and don't want to be wiped out by the first steep bit I come across!