How to jump a sportbike on the highway

# Thread: How to jump a sportbike on the highway

1. ## How to jump a sportbike on the highway

http://www.newson6.com/story/1505977...e-county-crash

The OHP says when Lewis' 2008 Honda CBR hit the ramp, it went airborne for 150 feet, then skidded another 222 feet.

2.

3. Damn, someone better call Ron Swanson to help get this fixed.

4. GFG........ i'm about to hit the states on my '08 1kRR..... dun wanna go airborn.... :P

5. Originally Posted by JoWii
GFG........ i'm about to hit the states on my '08 1kRR..... dun wanna go airborn.... :P
Paint your bike bright orange with confederate flag on the tank and you'll be just fine.

6. Could someone who remembers the formula (something that includes 9.8m/sec sqrd I recall) please calculate the speed at which a 650 lb object (bike & rider) must have been travelling to have sailed 150 ft from a 2 ft elevation? Bonus points for pictures that include dotted lines of travel.

7. Maybe he would have stuck the landing if he was on a motard? =)

8. Originally Posted by Stiffler's Mom
Could someone who remembers the formula (something that includes 9.8m/sec sqrd I recall) please calculate the speed at which a 650 lb object (bike & rider) must have been travelling to have sailed 150 ft from a 2 ft elevation? Bonus points for pictures that include dotted lines of travel.
Can't be done without a launch angle or velocity at takeoff (including direction), possibly more. Dropping off a 2 foot elevation would require a LOT more speed to cover the same distance as having a velocity in the upward direction as well as horizontal would for the same elevation point.

9. Thats good distance!!! on a dirt bike you got a big ramp and need 90+ MPH to do over 200 feet...Should a done a nac nac ...

10. Originally Posted by Stiffler's Mom
Could someone who remembers the formula (something that includes 9.8m/sec sqrd I recall) please calculate the speed at which a 650 lb object (bike & rider) must have been travelling to have sailed 150 ft from a 2 ft elevation? Bonus points for pictures that include dotted lines of travel.

11. 110 degrees... the sun isnt big enough in that diagram

12. Riding in 110 degree heat is like standing in front of 30 hair driers. Some days it's just too hot to ride.

13. You can get the launch angle from the picture in the article (the ramp is about 6 feet long and they said 2 feet high so about 20 degree angle), so solve for the air time in the horizontal direction (Vcos20= t/46meters) and in the vertical direction using a downward acceleration of 9.81m/s^2 (Vsin20t + 1/2(9.81)t^2) and equate them through the time variable. Then you will have the x and y components of initial velocity and just inverse tangent that ratio. It works out to be about what Chia drew.

14. Originally Posted by Bansal
Can't be done without a launch angle or velocity at takeoff (including direction), possibly more. Dropping off a 2 foot elevation would require a LOT more speed to cover the same distance as having a velocity in the upward direction as well as horizontal would for the same elevation point.
The ramp has a 13.3329...% grade

15. Originally Posted by Estefaz
You can get the launch angle from the picture in the article (the ramp is about 6 feet long and they said 2 feet high so about 20 degree angle), so solve for the air time in the horizontal direction (Vcos20= t/46meters) and in the vertical direction using a downward acceleration of 9.81m/s^2 (Vsin20t + 1/2(9.81)t^2) and equate them through the time variable. Then you will have the x and y components of initial velocity and just inverse tangent that ratio. It works out to be about what Chia drew.
The concrette sections are 15 feet long (as stated in the story)

16. Originally Posted by Estefaz
You can get the launch angle from the picture in the article (the ramp is about 6 feet long and they said 2 feet high so about 20 degree angle), so solve for the air time in the horizontal direction (Vcos20= t/46meters) and in the vertical direction using a downward acceleration of 9.81m/s^2 (Vsin20t + 1/2(9.81)t^2) and equate them through the time variable. Then you will have the x and y components of initial velocity and just inverse tangent that ratio. It works out to be about what Chia drew.

He wins.

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