Ok, I may be a simpleton. In many ways I definitely am. But if you have a power meter and a speedometer of some fashion, which your post suggests you do, why not just ride your bike a little bit harder than your Oly pace and see how many watts it takes to get to 25.5 mph*. You don't have to do this for very long, your power/speed ratio doesn't care how long you do it for.

*I assume you mean miles per hour and not miles per gallon as per the subject header, unless this is something to do with energy drink..

Good point. Answer:
1. I didn't think of it.
2. I'm lazy.

I hardly ever get outside on my tri bike in full aero getup, but I'll give it a try next time I do.

Thanks! That looks very useful, but I'm not sure how to read it, and will being at elevation ( about 5280) change it significantly?

the answer can vary as much as 50 watts or more for people of the same size and shape, when both are on tt bikes!

so, depends.

analyticcycling.com can spit out an answer, but you have to input a guess at your CdA and rolling resistance

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You want to ride ~40kph. So pick out that black dotted line. Now you have to make an assumption of your CdA (drag coefficient) this will vary based on your body size and equipment. You can use analytic cycling or BestBikeSplit with data from a past race to get a feel for it. With a CdA=0.25, you'd be looking at ~250 watts to maintain 40kph.

Power required is decreases linearly with altitude. So to make this chart apply, use the ratio of sea level air density to 5280 feet air density: 1.0467/1.225 or ~85%.

Whats your average RPM and what gear are you in the longest?

Example: ~95 rpm in 53x15 or 50/14 (~96 inch gear) to average between 24-27mph.

All the helpful answers are a bit more than you are bargaining for. The exact answer is obviously complicated, but roughly speaking, the necessary power scales as the cube of the speed. This means that if you want to go 10% faster you need ~30% more power. So in your case, a guess might be that you need ~255 W.

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I suspect that you'd need to average closer to 270w to get to 25.5mph, that's a couple km/h faster than 40.
I just did a similar trial to calculate my CdA on my new bike and slight position change.

-SD

I think the 255 is close but could be as high as 270. Somewhere in that range. Roughly, I get a minute per watt at IM distances and about 1 mph per 20 watts. This starts to fall apart above 40 kph though.....as harder to go 24 to 25 than 20 to 21....

I do agree that you are fairly lame for not figuring this out for yourself given that you have a PM.....but this is slowtwitch so your question is relatively reasonable too....

Although your sea level and 5280 speed at a given wattage will be quite different....

Dave

Brilliant plot! Left is riding on the hoods, no attempt at being aero (clothing, etc) on a road frame. Right is a pathologically good time trial position on a UCI legel bike...like full aero tuck descending (maybe).

Is this 22.8mph in a race? or steady state on flat pavement? IN a race there's a lot of "overhead" going and and out of T1/T2 that adds some time. Plus there's turns that limit speed as well at times. So to go faster, you have to go "even faster".

I'd say that the range would be 240-260W.... maybe 235-270 if you have a Stages PM depending on which leg is stronger.

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All these are nice estimates/speculations, but only the OP will be able to precisely answer his own question. All it takes is a couple of minutes holding 25.5 or 26 mph on a stretch of road and take the average power output from that.

This will be very course and condition dependent. Here are my stats from Church Creek this past June. 40k. Shiv, TriRig bits, Reynolds RZR92s, no cover. 6ft and 180lbs. 57:33 (25.85mph) on 281w avg and 282 normalized. Both were personal bests at the time. I've since gone slower (25.3) on a shorter, flatter course on 284/288.

hth

The exact answer is obviously complicated, but roughly speaking, the necessary power scales as the cube of the speed. This means that if you want to go 10% faster you need ~30% more power. So in your case, a guess might be that you need ~255 W.

The 2.7 power is a better approximation.

P2=P1*(V2/V1)^2.7

So in the OP's case we have P2= 193*(25.5/22.8)^2.7= 261W

I like simple solutions. Forget targeting a time/goal, you'll set yourself up for failure or sell yourself short. Determine how much time you can train and develope a plan to maximize you fitness within those time limitations. Get as strong as you can on the bike by raising your FTP. Work on experimenting with your position.

Why do I say this? Well I only increased my FTP by 6% over 7 months of IM focused training. But at the same time I set a goal of 2:15 for a 70.3. I went 2:17 in my first race, then 2:09 in my 2nd. My average power was almost identical in both races. So how did I do compared to my goal? Who gives a shit! I rode a f***** 2:09 and kicked ass that day. Of course I folded on the run... but nobody cares about that.

On the flip since in my only Olympic, hot and humid, swam hard, and coming off a hard training block my legs just weren't there. I still set a PR but missed my time goal. I took the performance in the full context of where I was in my training.

Get fit, figure out a target power and nail it.

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