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Silent pipes take lives
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:bzbatl
 

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OO=[][]=OO
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At the end of the day, we are discussing basic calculus.

A = dv/dt = d/dt(dx/dt) = d^2x/dt^2

t^2 is quite accurate. I can’t tell you how many times I had to derive those kinematic equations in college, I had a professor in Dynamics, IIRC, who insisted that every answer start with a=dv/dt, v=dx/dt etc
 

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A guy on a scruffy bike
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lets just use meters...

m/s/s = m*s/s = m
Bzzt. m/s/s does not equal m*s/s.

m/s/s = m * 1/s * 1/s = m/s^2

Position is meters: m
Velocity is change in position, meters per second: m/s
Acceleration is change in velocity, meters per second per second: m/s/s or m/s^2

So s/s is the same as s*s?
No. But s/s is not the same as /s/s, nor is s*s the same as /s*s.

Dividing by something is the same as multiplying by its inverse. So dividing by 2, for instance, is the same as multiplying by 1/2.

So meters divided by seconds is: m * 1/s = m/s.
And then m/s divided by seconds again is: m/s/s = (m/s)/s = m/s * 1/s = m/s*s = m/s^2. Same thing.

PhilB
 

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No more than 6 if it's a UJM. Eurobikes, a decent amount higher. If you go domestic you'll easily pull 10, sometimes twice that.
 

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Silent pipes take lives
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you're right, i was taking it as m/(s/s), i miss understood scissor's post
So did Phil, since I was addressing the intuitive aspect of the formula rather than the actual math.

But Phil, being a Vulcan, missed that.
 

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A guy on a scruffy bike
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So did Phil, since I was addressing the intuitive aspect of the formula rather than the actual math.

But Phil, being a Vulcan, missed that.
What the hell is "the intuitive aspect of the formula"? Yeah, I missed that alright. That's why we have math -- because intuition isn't a reliable method of actually figuring stuff out.

PhilB
 

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Track Junkie
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Bikes don't do "lateral" Gs, due to lean angles and gravity canceling out the effect. They do downward pull type Gs, like that of a fighter jet when banking for a turn. The feeling of riding a bike is more like flying a plane than driving a car.

DocMC | Motorcycle Doctor
 

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A guy on a scruffy bike
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Bikes don't do "lateral" Gs, due to lean angles and gravity canceling out the effect. They do downward pull type Gs, like that of a fighter jet when banking for a turn. The feeling of riding a bike is more like flying a plane than driving a car.

DocMC | Motorcycle Doctor
Not true. The lean angle is necessary to balance the bike WHILE doing the lateral Gs. If there were no lateral Gs involved, tires wouldn't be important in cornering. Your reference does not show that motorcycles don't have lateral Gs; it shows that if you tilt your accelerometer relative to the lateral direction, it doesn't measure lateral Gs anymore. Well, duh.

Note that their graph *does* show lateral Gs as measured by the GPS. (And, helps answer the question of this thread, by showing those lateral Gs peaking at about 1.5 Gs.)

PhilB
 

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There are lateral Gs involved, but when it comes to measuring lateral Gs for performance, not possible with tools like what are used in cars. Thats what I meant by don't do lateral Gs, sorry, didn't explain it well :)
 

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A guy on a scruffy bike
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There are lateral Gs involved, but when it comes to measuring lateral Gs for performance, not possible with tools like what are used in cars. Thats what I meant by don't do lateral Gs, sorry, didn't explain it well :)
You'd have to put it in a gimbal mount so the accelerometer doesn't lean with the bike. Or use the GPS.

PhilB
 

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A guy on a scruffy bike
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i think the results could be had with a 3 axis accelerometer and a bit of math.
Yes, if you had some way to measure lean angle in real time as well.

PhilB
 

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Sorry I know nothing about bikes so I could be way off but assuming the upright position of a bike is 0° if you then leaned it to 90° would you not just be laying it down
 

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It's kind of a non-sensical thread. Most people would pass out with 5 G's of acceleration force. Wiki This is why pilots wear G-suits to compress the legs and maintain blood flow to the brain to help avoid that. A really high G roller coaster is said to make 5G's. I think one of the earlier posts suggested maybe 2 G's of acceleration force, which sounds a lot more plausible w/o any actual measurements being done.

In terms of leaning at 90º, think of a high bank race track like Dallas Super Speedway. You're at 90º on the high bank, but you don't really steer around a track like that, you just ride/drive strait and go around. BTDT. It's an experience and you feel the increase in G forces, but it's not massive.
 
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