Ramp Slope worksheet

1. Slope ratio (rise : run): 1:12

1. Find the slope: turn the run into inches, then divide it by the rise to see how far the ramp runs for each inch it climbs

   $\frac{288\text{\hspace{0.17em}in}}{24\text{\hspace{0.17em}in}}$$=12$$=\text{slope}\text{1:12}$

2. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{288}^2+{24}^2}$$=288.9983\text{\hspace{0.17em}in}$

3. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{24}{288}$$=4.7636\text{\hspace{0.17em}°}$

4. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{24}{288}\times 100$$=\mathrm{8.3333\%}$

5. Each run may climb at most 24 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{24}{24}$$=1$$=\text{rounded up:}1\text{runs,}0\text{landings}$

6. Check against your 1:13 limit: that slope needs a run of the rise × 13

   $24\times 13$$=312\text{\hspace{0.17em}in}\text{, more than the run available}$

7. Subtract to see how much more run you would need

   $312-288$$=24\text{\hspace{0.17em}in}$

2. Minimum horizontal run: 18.33 ft

1. A 1:11 slope runs 11 inches along the ground for every inch it climbs, so multiply the rise by 11

   $20\times 11$$=220\text{\hspace{0.17em}in}$

2. Divide by 12 to turn inches into feet

   $\frac{220}{12}$$=18.3333\text{\hspace{0.17em}ft}$

3. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{220}^2+{20}^2}$$=220.9072\text{\hspace{0.17em}in}$

4. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{20}{220}$$=5.1944\text{\hspace{0.17em}°}$

5. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{20}{220}\times 100$$=\mathrm{9.0909\%}$

6. Each run may climb at most 32 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{20}{32}$$=0.625$$=\text{rounded up:}1\text{runs,}0\text{landings}$

3. Minimum horizontal run: 20 ft

1. A 1:12 slope runs 12 inches along the ground for every inch it climbs, so multiply the rise by 12

   $20\times 12$$=240\text{\hspace{0.17em}in}$

2. Divide by 12 to turn inches into feet

   $\frac{240}{12}$$=20\text{\hspace{0.17em}ft}$

3. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{240}^2+{20}^2}$$=240.8319\text{\hspace{0.17em}in}$

4. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{20}{240}$$=4.7636\text{\hspace{0.17em}°}$

5. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{20}{240}\times 100$$=\mathrm{8.3333\%}$

6. Each run may climb at most 30 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{20}{30}$$=0.6667$$=\text{rounded up:}1\text{runs,}0\text{landings}$

4. Slope ratio (rise : run): 1:8.47

1. Find the slope: turn the run into inches, then divide it by the rise to see how far the ramp runs for each inch it climbs

   $\frac{288\text{\hspace{0.17em}in}}{34\text{\hspace{0.17em}in}}$$=8.4706$$=\text{slope}\text{1:8.47}$

2. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{288}^2+{34}^2}$$=290\text{\hspace{0.17em}in}$

3. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{34}{288}$$=6.7329\text{\hspace{0.17em}°}$

4. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{34}{288}\times 100$$=\mathrm{11.8056\%}$

5. Each run may climb at most 34 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{34}{34}$$=1$$=\text{rounded up:}1\text{runs,}0\text{landings}$

6. Check against your 1:14 limit: that slope needs a run of the rise × 14

   $34\times 14$$=476\text{\hspace{0.17em}in}\text{, more than the run available}$

7. Subtract to see how much more run you would need

   $476-288$$=188\text{\hspace{0.17em}in}$

5. Minimum horizontal run: 15 ft

1. A 1:10 slope runs 10 inches along the ground for every inch it climbs, so multiply the rise by 10

   $18\times 10$$=180\text{\hspace{0.17em}in}$

2. Divide by 12 to turn inches into feet

   $\frac{180}{12}$$=15\text{\hspace{0.17em}ft}$

3. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{180}^2+{18}^2}$$=180.8978\text{\hspace{0.17em}in}$

4. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{18}{180}$$=5.7106\text{\hspace{0.17em}°}$

5. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{18}{180}\times 100$$=\mathrm{10\%}$

6. Each run may climb at most 28 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{18}{28}$$=0.6429$$=\text{rounded up:}1\text{runs,}0\text{landings}$

6. Slope ratio (rise : run): 1:12.71

1. Find the slope: turn the run into inches, then divide it by the rise to see how far the ramp runs for each inch it climbs

   $\frac{216\text{\hspace{0.17em}in}}{17\text{\hspace{0.17em}in}}$$=12.7059$$=\text{slope}\text{1:12.71}$

2. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{216}^2+{17}^2}$$=216.6679\text{\hspace{0.17em}in}$

3. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{17}{216}$$=4.5001\text{\hspace{0.17em}°}$

4. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{17}{216}\times 100$$=\mathrm{7.8704\%}$

5. Each run may climb at most 38 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{17}{38}$$=0.4474$$=\text{rounded up:}1\text{runs,}0\text{landings}$

6. Check against your 1:13 limit: that slope needs a run of the rise × 13

   $17\times 13$$=221\text{\hspace{0.17em}in}\text{, more than the run available}$

7. Subtract to see how much more run you would need

   $221-216$$=5\text{\hspace{0.17em}in}$

7. Minimum horizontal run: 36.83 ft

1. A 1:17 slope runs 17 inches along the ground for every inch it climbs, so multiply the rise by 17

   $26\times 17$$=442\text{\hspace{0.17em}in}$

2. Divide by 12 to turn inches into feet

   $\frac{442}{12}$$=36.8333\text{\hspace{0.17em}ft}$

3. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{442}^2+{26}^2}$$=442.764\text{\hspace{0.17em}in}$

4. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{26}{442}$$=3.3665\text{\hspace{0.17em}°}$

5. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{26}{442}\times 100$$=\mathrm{5.8824\%}$

6. Each run may climb at most 22 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{26}{22}$$=1.1818$$=\text{rounded up:}2\text{runs,}1\text{landings}$

7. Add the landings between runs to get the overall length on the ground (the top and bottom landings are extra)

   $442+1\times 60$$=502\text{\hspace{0.17em}in}$$=41.8333\text{\hspace{0.17em}ft}$

8. Slope ratio (rise : run): 1:10

1. Find the slope: turn the run into inches, then divide it by the rise to see how far the ramp runs for each inch it climbs

   $\frac{240\text{\hspace{0.17em}in}}{24\text{\hspace{0.17em}in}}$$=10$$=\text{slope}\text{1:10}$

2. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{240}^2+{24}^2}$$=241.197\text{\hspace{0.17em}in}$

3. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{24}{240}$$=5.7106\text{\hspace{0.17em}°}$

4. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{24}{240}\times 100$$=\mathrm{10\%}$

5. Each run may climb at most 27 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{24}{27}$$=0.8889$$=\text{rounded up:}1\text{runs,}0\text{landings}$

6. Check against your 1:9 limit: that slope needs a run of the rise × 9

   $24\times 9$$=216\text{\hspace{0.17em}in}\text{, no more than the run available}$

9. Slope ratio (rise : run): 1:10

1. Find the slope: turn the run into inches, then divide it by the rise to see how far the ramp runs for each inch it climbs

   $\frac{240\text{\hspace{0.17em}in}}{24\text{\hspace{0.17em}in}}$$=10$$=\text{slope}\text{1:10}$

2. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{240}^2+{24}^2}$$=241.197\text{\hspace{0.17em}in}$

3. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{24}{240}$$=5.7106\text{\hspace{0.17em}°}$

4. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{24}{240}\times 100$$=\mathrm{10\%}$

5. Each run may climb at most 32 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{24}{32}$$=0.75$$=\text{rounded up:}1\text{runs,}0\text{landings}$

6. Check against your 1:15 limit: that slope needs a run of the rise × 15

   $24\times 15$$=360\text{\hspace{0.17em}in}\text{, more than the run available}$

7. Subtract to see how much more run you would need

   $360-240$$=120\text{\hspace{0.17em}in}$

10. Slope ratio (rise : run): 1:8.31

1. Find the slope: turn the run into inches, then divide it by the rise to see how far the ramp runs for each inch it climbs

   $\frac{216\text{\hspace{0.17em}in}}{26\text{\hspace{0.17em}in}}$$=8.3077$$=\text{slope}\text{1:8.31}$

2. The run, rise and ramp surface form a right triangle, so the Pythagorean theorem gives the length of the sloped surface

   $\sqrt{{216}^2+{26}^2}$$=217.5592\text{\hspace{0.17em}in}$

3. Find the angle: arctan turns rise ÷ run into degrees

   $\text{arctan}\frac{26}{216}$$=6.8637\text{\hspace{0.17em}°}$

4. Find the grade: the slope as a percent, or the rise for every 100 of run

   $\frac{26}{216}\times 100$$=\mathrm{12.037\%}$

5. Each run may climb at most 26 in, so divide the rise by that and round up. A level landing goes between runs

   $\frac{26}{26}$$=1$$=\text{rounded up:}1\text{runs,}0\text{landings}$

6. Check against your 1:10 limit: that slope needs a run of the rise × 10

   $26\times 10$$=260\text{\hspace{0.17em}in}\text{, more than the run available}$

7. Subtract to see how much more run you would need

   $260-216$$=44\text{\hspace{0.17em}in}$