What is the ballpark conversion from grams of drag to time saved in a IM bike at MOP speeds
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50 grams of drag (measured at 30mph)—> ~0.5 secs per kilometer (at normal biking speeds)
112 miles = 180 kilometers
180 kilometers *.5 = 1.5 minutes per ironman
assuming it is flat and you are not a drafting bastards
if you remember the 50 → 0.5s/k rule of thumb, google can do the rest of the conversions for you
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i thought it was 45g= .5s= 5 watts= cda =.005
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Thanks,
Also, I know this depends a lot on wind direction, but what are the typical yaws faced by a rider.
On the drag charts, what part of the chart is more important, 0 degrees, 5, 10, 15, 20, etc.
Obviously if a piece of equipment is better across the drag sweep it is better, but sometimes some are better at some yaw angles than others.
Any rules of thumb when evaluating data.
i thought it was 45g= .5s= 5 watts= cda =.005
45g is slightly more accurate since the original *rough “*Rule Of Thumb” (ROT) was 0.1 lbs of drag (measured at 30mph) ~= 5 W (at race speeds) ~= 0.5 s/km ~= .005 m^ of CdA
(0.1 lbs = 45.4g)
However, making it 50g still keeps it useful as a rough ROT (which is the whole point of it) and makes it so the only number you have to remember is 5 
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depends on how windy it is and how fast you are =)
if your races are longer ones in exposed areas, you will be going slower, winds tend to pick up as the day goes on, yaw tends to be higher
if your races a short ones in neighborhoods, you are going faster, earlier in the morning, with shielding from the wind. lower yaw angles.
if you are doing a TT in fort davis texas in the summer, yaw angles are 90 degrees
Thanks,
Also, I know this depends a lot on wind direction, but what are the typical yaws faced by a rider.
On the drag charts, what part of the chart is more important, 0 degrees, 5, 10, 15, 20, etc.
Obviously if a piece of equipment is better across the drag sweep it is better, but sometimes some are better at some yaw angles than others.
Any rules of thumb when evaluating data.
50 grams of drag (measured at 30mph)—> ~0.5 secs per kilometer (at normal biking speeds)
112 miles = 180 kilometers
180 kilometers *.5 = 1.5 minutes per ironman
assuming it is flat and you are not a drafting bastards
if you remember the 50 → 0.5s/k rule of thumb, google can do the rest of the conversions for you
Is there a direct correlation? Does 50 grams of drag at 20mph (2/3) = 0.33 seconds per kilometer (2/3)? I thought I heard it is not a direct correlation and therefore makes the “rule of thumb” useless? I defer.
the rule of thumb lets you convert grams of drag measured at 30mph (because wind tunnels almost always measure at 30mph, so this is the data you will tend to see)
into time saved at normal biking speeds.
the TIME saved over a fixed distance is relatively constant over normal cycling speeds, by the way
remember that this is a rough rule of them, useful for getting a grip on the order of magnitude of time being saved - “am I saving minutes? seconds? nearly nothing?”
etc
I never understood why drag was measured in grams…why not Newtons?
grams~mass
newtons~force
right?
Thanks,
Also, I know this depends a lot on wind direction, but what are the typical yaws faced by a rider.
On the drag charts, what part of the chart is more important, 0 degrees, 5, 10, 15, 20, etc.
Obviously if a piece of equipment is better across the drag sweep it is better, but sometimes some are better at some yaw angles than others.
Any rules of thumb when evaluating data.
A 45 degree, 10mph headwind/crosswind, would net a 16 degree yaw angle at 25mph. Go the other way now at 35mph and the yaw angle is now 12 degrees. 10mph is a pretty strong wind. Cut that wind in half to 5mph and those yaw values are now halved (8 and 6).
The new FC wheels and Hed wheels get very fast at the higher yaw values, I heard a very popular bike that has a very popular thread on ST is very fast at the higher yaws do to the tube shapes. Even a certain hydration company I know…hint …hint, gets very fast at higher yaw values do to side force (lift) in greater than 90 degrees of the opposing force of drag.
I never understood why drag was measured in grams…why not Newtons?
grams~mass
newtons~force
right?
i hereby make a motion that whenever testers talk about drag they report their values in cda m
I also suspect that many companies would have some extremely small time savings at say 20-24 mph. And when we read the advertisements we kindda forget that we are not able to do 30mph. and i have no idea what the mathemathical formula from 30 ----> my 23 mph is.
Putting it very simply, power needed to overcome wind resistance grows as a cube. If you need 1 unit of power at 30 units of speed, you will need 0.00003704 * (speed cubed) units of power at a different speed. Talking about something that will require 1 extra unit of power at 30mph, that’ll be 0.512 units of power needed at 24mph.
Showing my work:
f = cx^3 ← c is an unknown coefficient
1 = c(30^3) ← because we know we need one unit at 30mph
1 / 30^3 = c ← divide both sides by x^3
3.704e-5 = c
Edited for correct wording (my calculations were for power, but my original post said force). See Tom A’s post below.
Putting it very simply, wind resistance grows as a cube. If you have 1 unit of resistance at 30 units of speed, you will have 0.00003704 * (speed in mph cubed) units of resistance at a different speed. For 24 units of speed, that’ll be 0.512 units of resistance.
Showing my work:
f = cx^3 ← c is an unknown coefficient
1 = c(30^3) ← because we know we have one unit at 30mph
1 / 30^3 = c ← divide both sides by x^3
3.704e-5 = c
Actually, no.
Drag force varies with the square of the velocity. Power varies with the cube.
If you want to know the drag force at 24 mph if you know it at 30 mph, then just take the ratio of the squares, or 24^2 / 30^2 = .64, or 64% of the 30 mph value.
That said, the ROT is geared towards “real world” equivalents in the range of ~22 - 28 mph for data taken at 30mph of tunnel speed. I find it actually works fairly well under various situations. For example, I have a friend who has nearly the exact same drag as myself, but can put out significantly more power. If I know either how much faster he went than I did, OR if I know what power he averaged, I can usually guess fairly accurately the other value (i.e. guess watts if knowning time difference). I’ve also found that it works pretty accurately for me for figuring out time savings and comparing them to actual performance (given conditions and equipment that are similar).
Pretty much every 10W of output for me is “worth” 1s/km…consistently.
I also suspect that many companies would have some extremely small time savings at say 20-24 mph. And when we read the advertisements we kindda forget that we are not able to do 30mph. and i have no idea what the mathemathical formula from 30 ----> my 23 mph is.
It doesn’t work that way. Even though the drag force is decreased by the ratio of the squares of the velocities (i.e. 23^2/ 30^2 = .59), you race over a fixed distance (not time) and so you are “on course” longer and so your absolute time savings are actually greater than someone who is faster*.*
It’s counterintuitive, I know…but the Cervelo folks covered it fairly well in one of their tech presentations…however, now that I look, I can’t seem to find it…
I also suspect that many companies would have some extremely small time savings at say 20-24 mph. And when we read the advertisements we kindda forget that we are not able to do 30mph. and i have no idea what the mathemathical formula from 30 ----> my 23 mph is.
sigh
this is one of those things
like wheel inertia
where knowing a little physics is worse than knowing none at all =)
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Yes its just like that
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Whoops - what I meant was power. Should have read “power needed to overcome wind resistance”. Will edit the post.
where knowing a little physics is worse than knowing none at all =)
It turns out using the wrong word in my post makes what I say nonsensical and incorrect. Who would have thunk it?
Here’s the calculation for wind resistance (which grows as a square of your velocity). If you know something causes 1 gram of drag at 30mph, then it should cause ((x)^2)/900 grams of drag at x mph.
f = cx^2 ← x = speed in mph
1 = c * 30^2
1/900 = c
f = (x^2)/900
Why does this type of banter not happen in any of the Fantasy Football forums?