If your aim is to lose weight, cutting caloric intake will likely be more helpful that riding fast.

While I wasn't thinking specifically about wind resistance when mentioning "efficiencies," you are correct that the faster speed will require more force. How much of your total power goes toward resisting the wind versus everything else (friction of the chain, the actual work required to move ___ lbs over a certain distance, etc.) would be interesting to know.

It's been mentioned already here. You can also fool around with the online power calculator Bicycle Speed (Velocity) And Power Calculator

Over a fixed distance, riding harder will require more work and therefore burn more calories even though it will take you less time.

I have a flat route that I ride often. It's 10.8 miles. I compared a hard ride against a mellow one.

Hard ride: 27:21, 23.8mph, 295w, 484kj (https://www.strava.com/activities/57...ts/14048151684)
Mellow ride: 34:13, 18.4mph, 172w, 366kj (https://www.strava.com/activities/72...ts/17797804990)

There's a difference, but it's probably not as great as you might expect. The difference between an on-the-rivet TT effort and a smell-the-daisies recovery ride is 118 calories over almost 11 miles. That's like two Oreos. For 3.7 miles, it's going to be much smaller.

Covering the same distance in less time on a bike burns more calories because you must overcome air resistance, which is proportional to your speed.

It takes more "force" (actually more power) to cover the same distance in less time.

whatever gets you there...

(doing it burns more kcal than thinking about doing it)

Talking to cats burns a hell of a lot of calories.

Correction.

Air resistance (drag) is proportional to the square of speed.

Correction:


Air resistance is proportional to the negative half of the drag coefficient, cross sectional area, and air density, and grows exponentially with velocity squared.

Moving on a bike at a faster speed will burn more calories per minute but maybe not overall any more than at a slower speed. But there is something to be said about increased wind resistance effort and then, after the ride is over, your metabolism still stays elevated.

Even though it is a short ride, I would push myself to ride it as fast as I am capable of and log it on an app like Strava about once a week to chart my progress. Racing myself is a good motivator but yielding health benefits makes it that much better.

You should have left well enough alone.
You can't seriously believe that air resistance pushes a rider ahead (negative resistance is propulsion).
You can't seriously believe that drag is proportional to e raised to the power of v squared (grows exponentially with velocity squared).

if we're getting geeky and picky here:

Air resistance is proportional to the negative half of the drag coefficient, cross sectional area, and air density, and grows exponentially with in proportion to velocity the apparent wind speed squared.



(sign is unnecessary here, you don't need the constant when saying "proportional", it's not exponential, and velocity in this question is the bike's ground speed)

toss the bike and walk

By your logic, a steeper hill is harder than flat land. Has it occurred to you that a bicycle has a freewheel and can go as fast it wants without you doing any work? It's that freewheel that makes descents and tailwinds effortless.

Take your bike out in a gale that would knock you off your feet. Turn it downwind and enjoy a free ride at 30+ mph.

I've heard rumors about these "tail winds" but every time I find a wind like that, it's always hitting me in the face.

Not if it's a completely downhill ride and you're gaining speed by applying less brake pressure.

Negative - I'm referencing the formula itself without consideration of the object experiencing the drag. The formula can, and often is, broken down into simpler iterations for ease of discussion when certain variables like drag coefficient and air density are given for an entire problem set, or not considered at all for the sake of just making a rough estimate. I.e., you missed the point here entirely.

From "Theory of Wing Sections" by Abbot and von Doenhoff:

Drag = 1/2 roe V^2 S Cd where roe = the mass density of air, V= velocity, S = Area (wing area for this treatise), Cd = coefficient of drag

Drag (sir resistance) grows with velocity squared and is proportional to area, air density and coefficient of drag, ie the drag characteristics of that shape.

I'm just a dumb engineer trained in fluid mechanics. I have never heard of the 'negative half of" anything, nor that air resistance grows exponentially with velocity. Maybe I was asleep (for six years) or perhaps my professors were not the best. Granted, the reference I gave, although once considered the bible on wing sections and the reference from which many airplanes were designed (and flew), was published in 1949 and probably quite wrong by now.

Ben

I am referencing quadratic velocity dependence for large objects moving through a fluid...this has also escaped previous commentors though this is not surprising.

To say that air resistance grows exponentially with velocity is just another way to say that it is proportional to the square of velocity - which we can say because the exponent in question is on an identified variable - and is more a reference to the graphed function rather than the equation itself. The plots in a graph of the axes velocity and resistance do not rise in a linear fashion, they rise with an exponential slope.

It's a convoluted derivation which is simplified for the sake of rough estimation, as I posted above. The point here is to nit-pick the nit-picker by complicating the formulation for the sake of being a dick, though ironically those who responded to this have instead simplified the formula to a more basic estimation of air drag. Makes me chuckle.

That word, exponentially, I don't think it means what you think it means.

Okay let's do some basic algebra. Draw a graph with velocity and air resistance each on a respective axis - it doesn't even matter which goes where. Now plot some arbitrary points using the air resistance formula.


It should look something like this:


https://www.everythingmaths.co.za/sci...fffb562e27.png


Get it yet? My guess is that you did not read and/or understand the whole of the comment to which you are replying. Try a re-read.

If you're going to nit pick it's confusing when you use mathematical terms incorrectly.

It looks parabolic. Because it IS parabolic, not exponential.

You know the first rule of holes - when you find yourself at the bottom of a hole, stop digging.

I love when I get to write that (probably too much).