Jean-Marie Wampers
Jean-Marie Wampers (b. Uccle, 7 April 1959) is a former professional road racing cyclist from Belgium. He was a professional between 1981 and 1992, achieving his greatest triumph when he won Paris–Roubaix in 1989.
 | |
| Personal information | |
|---|---|
| Full name | Jean-Marie Wampers |
| Born | 7 April 1959 Belgium |
| Team information | |
| Current team | Retired |
| Discipline | Road |
| Role | Rider |
| Professional teams | |
| 1981 | Santini-Sella Italia |
| 1982 | GIS-Olmo |
| 1983–1984 | Europ Shop-Splendor |
| 1985–1987 | Hitachi |
| 1988–1991 | Panasonic |
| 1992 | Colstrop |
| Major wins | |
| Paris–Roubaix (1989) | |
Major results
- 1984
- Druivenkoers Overijse
- 1985
- Nationale Sluitingsprijs
- 1986
- 1st, Stage 4, Four Days of Dunkirk
- Rund um den Henninger-Turm
- 1989
- GP de l'Escaut-Schoten
- Scheldeprijs
- Paris–Roubaix
- Rund um den Henninger-Turm
gollark: Take two of the points, subtract one's Y coordinate from the other one's Y coordinate, and do the same for the X coordinates, and divide the difference in Y coords by the difference in X coords.
gollark: <@379441093558927391> I'm assuming that what you want to do is find the equation of the straight line going through those points. So to do that you need the gradient.
gollark: <@379441093558927391>
gollark: Once you have the gradient you just need to work out the y intercept, so put that into the straight line equation (y = mx + c), substitute in the x and y from one of the points, solve for c (y intercept), and you're done.
gollark: =tex m = \frac{y_2-y_1}{x_2-x_1}
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.