Can't break 40mph downhill

[Balderick]

I had a guy (friend of a friend) tell me over the weekend that he did 52 mph on his MTB. I was thinking to myself "Well that's B.S." Not that it's not possible, but if he could really do it, he'd be riding as a NOBRA pro (which he isn't).....

L

With slicks?

 

[wiredued]

I just went down the same mountain road I saw the race at for the first time and hit 44mph Now I can't break 45mph downhill I think this is the hill to try it on should I lower the handlebars? Should I get rid of the trunk bag or does the extra cargo help downhill?

 

[DannoXYZ]

I just went down the same mountain road I saw the race at for the first time and hit 44mph Now I can't break 45mph downhill I think this is the hill to try it on should I lower the handlebars? Should I get rid of the trunk bag or does the extra cargo help downhill?

No need to lower the handlebars, just get your head lower, bend your elbows and tuck them in. Keep the trunk bag, load it up with 100lbs of lead-weights or rocks and you'll add another 5mph to that downhill. Be sure to start braking EARLY to account for the added weight...

 

[wiredued]

That sounds like a good idea ride up to the top of a mountain then fill the trunk bag with rocks for the ride down.

No need to lower the handlebars, just get your head lower, bend your elbows and tuck them in. Keep the trunk bag, load it up with 100lbs of lead-weights or rocks and you'll add another 5mph to that downhill. Be sure to start braking EARLY to account for the added weight...

 

[aris]

O.K. just to clear some of this physics stuff up im curerntly in the middle of an AP Physics class. First off E(sigma)F=Ma so! the sum of the forces acting on an object is proportional to the mass times the acceleration. As i see it there are four forces acting on the rider: air resistance, gravity, friction, and rotating. P.E.=mgh the potentail engergy of a rider is equal to the his mass times gravity times height. So in a perfect situation at the bottom of the hill his velocity will be the square root of twice gravity times his original height [sqrt(2gh)=v] because k.e. gained will be equal to p.e. lost and k.e.=(1/2)mv^2. Taking into account the external forces gravity : f=ma a=9.8 m/s^2 ; rotational (the key here!!) F=ma (like always) but acceleration is not only a change in speed it is a change in direction, so your wheels are always accelerating!! thusly force = mass times the centripital acceleration or F=m(v^2/r). So the rotaional force of the wheel has a quadratic relationship with speed. The force of friction does decease with speed but minimal, the only diffence seen is between the static coffecient of friction and the kenetic (moving) coffecient of friction. Then air resistance doubles every mph faster. SO! EF=ma == 9.8m[sin x] -(.5mv^2)+(Force of friction)+Air resistance=0 because there is a point where the force of gravity is at equilbrium with the forces acting on. Sin x, x is the degree of the decent in degress, you are not free falling there fore you need to find the total decent distance not the length of the hill. I hope this clears up all the nonsense physics people were tryign to explain

 

[Peka]

aris spouts more physics than I've forgotten....


Lol, just pedal faster!!

BTW - This 73kg (160lb) rider hit 72.2kmh (45mph) yesterday on my 16kg(36lb) MTB

 

[wiredued]

With this information plus a $1.25 I could buy a cup of coffee. Aris how much does your physics book wiegh it might help my down hill speed?

O.K. just to clear some of this physics stuff up im curerntly in the middle of an AP Physics class. First off E(sigma)F=Ma so! the sum of the forces acting on an object is proportional to the mass times the acceleration. As i see it there are four forces acting on the rider: air resistance, gravity, friction, and rotating. P.E.=mgh the potentail engergy of a rider is equal to the his mass times gravity times height. So in a perfect situation at the bottom of the hill his velocity will be the square root of twice gravity times his original height [sqrt(2gh)=v] because k.e. gained will be equal to p.e. lost and k.e.=(1/2)mv^2. Taking into account the external forces gravity : f=ma a=9.8 m/s^2 ; rotational (the key here!!) F=ma (like always) but acceleration is not only a change in speed it is a change in direction, so your wheels are always accelerating!! thusly force = mass times the centripital acceleration or F=m(v^2/r). So the rotaional force of the wheel has a quadratic relationship with speed. The force of friction does decease with speed but minimal, the only diffence seen is between the static coffecient of friction and the kenetic (moving) coffecient of friction. Then air resistance doubles every mph faster. SO! EF=ma == 9.8m[sin x] -(.5mv^2)+(Force of friction)+Air resistance=0 because there is a point where the force of gravity is at equilbrium with the forces acting on. Sin x, x is the degree of the decent in degress, you are not free falling there fore you need to find the total decent distance not the length of the hill. I hope this clears up all the nonsense physics people were tryign to explain